Mathematics Tricks & Tips of Sequences and Series For NDA

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Tricks & Tips of Sequences and Series For NDA

Problem On AP

`T_n= a + (n - 1)d`

`S_n=n/2[2a+(n-1)d] = n/2(a+l)`

Short trick : Assume term in AP 1,2,3,4 as requirement

Q 2713780649

If the sum of m terms of an AP is n and the sum of `n` terms is `m`, then the sum of `(m + n)` terms is
NDA Paper 1 2017

(A)

`mn`

(B)

`m+n`

(C)

`2(m+n)`

(D)

`-(m+n)`

Solution:

For pure mathematicians detailed solution is given below

Given , sum of `m` terms `=n`

`=> S_m=n`

`=> 2am+m (m-1)d=2n`..............(1)

sum of `n` terms `=m`

`=> S_n =m`

`=> n/2 [ 2a +(n-1)d]=m`

`2an+n(n-1) d = 2m`...........(2)

subtracting equation (2) from equation (1) , we get

`2am +m(m-1)d -[2an+n(n-1)d] = 2n-2n`

`=> 2a (m-n)+d [m(m-1) -n (n-1)] = 2(n-m)`

`=>2a(m-n)+d[m^2-m-n^2+n] =-2 (m-n)`

`=> 2a(m-n) +d(m+n)(m-n) -(m-n)]=-2 (m-n)`

`=> 2a +(m+n-1)=-2` [Taking `(m-n)` common]

`2a+(m+n-1)d =-2`................(3)

Now, `S_(m+n) =((m+n)/2) [2a +(m+n-1)d]`

`=((m+n)/2) (-2)` [using (3)]

Short cut Method

Please check video for explanation.

Let `m=1, n=2` and series be `a, a+d +......`

`=> a=2` and `a+a+d =1=> d=-3`

Sum of `(1+2) ` terms `=` Sum of series `= (a+a+d + a+2d)`

`=3 (a+d) = 3 (2-3) =-3 =-(1+2)`

`=> ` option `4` is correct.

Correct Answer is `=>` (D) `-(m+n)`

Q 2317734689

If the numbers `n - 3, 4n -2, 5n + 1` are in `AP` then
what is the value of `n`?
NDA Paper 1 2013

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Given that, `(n- 3), (4n-2), (5n+ 1)` are in `AP`.

`:. (4n-2) - (n - 3) = (5n+ 1) - (4n - 2)`

`=> 3n + 1 = n + 3 => 3n - n = 3 - 1`

`=> 2n = 2`

`:. n = 1`

Correct Answer is `=>` (A) `1`

Q 2763180945

The fifth term of an AP of `n` terms, whose sum is `n^2 - 2n`, is
NDA Paper 1 2017

(A)

`5`

(B)

`7`

(C)

`8`

(D)

`15`

Solution:

`S_n= n^2- 2n`

`S_1 =a =-1`

`S_2 =0`

`n/2 (2a +(2-1)d=0`

`2(-1) + d=0`

`d=2`

`5^(th)` term `a+ 4d = -1 + 8 =7`

Correct Answer is `=>` (B) `7`

Q 2783180947

The sum of all the two - digit odd numbers is
NDA Paper 1 2017

(A)

`2475`

(B)

`2530`

(C)

`4905`

(D)

`5049`

Solution:

`11, 13, ...........99`

`s= n/2 (a+l)` ` \ \ \ \ \ \ \ ` `[tt ( ( l= a+ (n+1)d) , (99=11+(n-1)2) , (n=45) ) ]`

`=45/2 (11+99)`

`=2475`

Correct Answer is `=>` (A) `2475`

Q 2741756623

The interior angles of a polygon of n sides are in AP. The smallest angle is 120° and the common difference is 5°.
How many possible values can n have ?
NDA Paper 1 2016

(A)

one

(B)

two

(C)

three

(D)

Infinitely many

Solution:

`d= 5^o > 1`

smallest angle `=a =120`

`AP` will be

`120, 125,130.....`

Now, sum of interior angles of polygon

`= (n-2) 180`

`n/2 [2* 120 + (n-1) 5] = (n-2) * 180`

`5n^2 -125 n + 720=0`

`n^2 -25 n + 144 = 0`

`n= 9` or `n=16`

Correct Answer is `=>` (D) Infinitely many

Q 2251212124

If the `nth` term of an `AP` is `(3+n)/4`, then the sum of first `105`
terms is
NDA Paper 1 2015

(A)

`270`

(B)

`735`

(C)

`1409`

(D)

`1470`

Solution:

We have `t_n = (3+n)/4`

which is an `AP` whose first term = `1`

and common difference `= 1/4`

`:. S_(105) = (105)/2 [ 2 + 104 xx1/4 ] = (105)/2 xx 28`

` = 105 xx 14= 1470`

Correct Answer is `=>` (D) `1470`

Q 2231812722

What is the sum of `n` terms of the series
` sqrt(2) + sqrt(8) + sqrt(18) + sqrt(32) +...?`
NDA Paper 1 2015

(A)

`(n(n -1))/ sqrt(2)`

(B)

`sqrt(2)(n -1)`

(C)

`(n(n +1))/ sqrt(2)`

(D)

`(n(n -1))/ (2)`

Solution:

We have, ` sqrt(2) + 2 sqrt(2) + 3sqrt(2) + 4sqrt(2) + ....`

which is an `AP`.

Here, `a = sqrt(2)` and `d = sqrt(2)`

`∵ S_n = n/2 [2a + (n- 1)d]`

`:. S_n = n/2 [2sqrt(2) + (n - 1) sqrt(2)]`

` =n/2 [ 2 sqrt(2) + sqrt(2)n - sqrt(2)]`

`= n/2 [sqrt(2)n + sqrt(2)] = n/2 sqrt(2) (n + 1) = n/sqrt(2) (n + 1)`

Correct Answer is `=>` (C) `(n(n +1))/ sqrt(2)`

Q 2357267184

In an `AP`, if the `m`th term is `1//n` and `n`th term is
`1//m`, then what is its (`mn` )th term?
NDA Paper 1 2009

(A)

`1//(mn)`

(B)

`m//n`

(C)

`n//m`

(D)

`1`

Solution:

Given, `T_m = a + (m - 1)d`

`=> 1/n = a+ (m -1)d` ........(i)

and `T_n = a + (n - 1)d`

`=> 1/m = a+ (n -1)d` ....(ii)

On solving Eqs. (i) and (ii), we get

`a = d = 1/(mn)`

`:. T_(mn) = 1/(mn) + (mn - 1) 1/(mn) = 1`

Correct Answer is `=>` (D) `1`

Q 2522601531

Six numbers are in `AP` such that their sum is `3`. The first term is `4` times the third term. Then, the fifth term is
WBJEE 2012

(A)

`-15`

(B)

`-3`

(C)

`9`

(D)

`-4`

Solution:

Let the numbers be

`a - 5d, a - 3d, a - d, a + d, a + 3d, a + 5d`

`:. a - 5d +a- 3d+ a-d+ a+ d +a`

`+3d + a + 5d = 3`

( ∵ Sum = 3)

`=> 6a = 3 => a = 1/2`

Also, given `T_1 = 4T_3` , where `T_1, T_3` are

respectively, first and third terms of AP.

`=> a- 5d = 4 (a - d)`

`=> d = -3a = - 3/2`

`:.` The fifth term

`a + 3d = 1/2 + 3 (-3/2) = 1/2 - 9/2 = -4`

Correct Answer is `=>` (D) `-4`

Q 2231812722

What is the sum of `n` terms of the series
` sqrt(2) + sqrt(8) + sqrt(18) + sqrt(32) +...?`
NDA Paper 1 2015

(A)

`(n(n -1))/ sqrt(2)`

(B)

`sqrt(2)(n -1)`

(C)

`(n(n +1))/ sqrt(2)`

(D)

`(n(n -1))/ (2)`

Correct Answer is `=>` (C) `(n(n +1))/ sqrt(2)`

Q 2344134053

The sum of the series `1/2 + 1/3 + 1/6 + …` to `9`
terms is :
BITSAT Mock

(A)

` - 1/2`

(B)

`- 5/6`

(C)

`- 3/2`

(D)

`1`

Solution:

The series

`1/2 + 1/3 + 1/6 + … + 9` terms is in A.P.

The sum `S_n` of `n` terms of an `A.P`. with

first term ‘a’ and common difference ‘d’

is given by `S_n = n/2 [2 a + (n − 1) d]`

Here `n = 9, a = 1/2`,

`d = 1/3 − 1/2 = −1/6`

`∴ S_n = 9/2 [2 . 1/2 + (9 − 1) . −1/6]`

`= 9/2 [1 + 8 × (−1)/6 ] = 9/2 [1 − 4/3]`

`= 9/2 × (−1)/3 = (−3)/2`

Correct Answer is `=>` (C) `- 3/2`

Q 2332101032

If in an A.P. `a_1, a_2, a_3, ...; a_7 = 9`, then

`a_1a_2a_7` is least when the common difference is
BITSAT Mock

(A)

`23/20`

(B)

`13/22`

(C)

`43/20`

(D)

`33/20`

Solution:

`a_7 = a + 6d = 9 => a = 9 - 6d`

(where `a` is the first term and `d` is
the common difference of the A.P.)

`:. a_1a_2 a_7 = 9(9 - 6d)(9 - 6d + d)`

`= 9(81 - 99d + 30d^2)`

`=270 [ (d- 33/20)^2 - 9/400]`

which is least, when `d = 33/20`

Correct Answer is `=>` (D) `33/20`

Q 2387278187

The difference between the nth term and `(n -1)` th
term of a sequence is independent of `n`. Then, the
sequence follows which one of the following?
NDA Paper 1 2008

(A)

(B)

(C)

(D)

None of these

Solution:

Let `n`th term be `an + b`.

So, `(n - 1)` th term is `a (n - 1) + b`.

Difference between these terms `= an + b - an + a - b = a`.

Hence, the sequence is in `AP` for which difference between the

`n`th term and `(n - 1)` th term is independent of `n`.

Correct Answer is `=>` (A) AP

Q 2327178081

After paying `30` out of `40` instalments of a debt of
`Rs. 3600`, one-third of the debt is unpaid. If the
instalments are forming an arithmetic series, then
what is the first instalment?
NDA Paper 1 2008

(A)

`Rs. 50`

(B)

`Rs. 51`

(C)

`Rs. 105`

(D)

`Rs. 110`

Solution:

Let first instalment be `Rs. x` and difference of

consecutive instalments be `Rs. d`.

`:. ( 30)/2 (2x + 29d) = (3600 xx 2)/3 => 2x + 29d = (2400)/(15)`

`=> 2x + 29d = 160` ... (i)

and ` (40)/2 [2x + 39d] = 3600 => 2x + 39d = 180` ... (ii)

On solving Eqs. (i) and (ii), we get

`x =51` and `d = 2`

`:.` First instalment `= Rs. 51`

Correct Answer is `=>` (B) `Rs. 51`

Q 2357756684

What is the sum of all natural numbers between
`200` and `400` which are divisible by `7`?
NDA Paper 1 2010

(A)

`6729`

(B)

`8712`

(C)

`8729`

(D)

`9276`

Solution:

The numbers between `200` and `400` which are divisible

by `7`, are

`203 ,210 ,217, ... ,399`

Now, let the number of terms be `n`.

`:. 399 = 203 + (n - 1)7`

`=> (196)/7 =(n-1) => n = 29`

Thus, required sum `= (29)/2 [203 + 399] = ( 29 xx 602)/2 = 8729`

Correct Answer is `=>` (C) `8729`

Q 2347734683

The sum of the first `5` terms and the sum of the
first `10` terms of an `AP` are same. Which one of the
following is the correct statement?
NDA Paper 1 2013

(A)

The first terms must be negative

(B)

The common difference must be negative

(C)

Either the first term or the common difference is negative but not both

(D)

Both the first term and the common difference are negative

Solution:

Let a be the first term and d be the common difference

of an AP.

`S_5 = S_(10)` (by condition)

`=> 5/2 (2a + 4d) = (10)/2 (2a + 9d)`

`=> a + 2d = 2a + 9d => a + 7d = 0`

`:. a = - 7d`

Let first `5` terms of an AP are `a - 2d, a-d , a , a + d` and `a + 2d`.

`=> - 9d, - 8d, - 7d, - 6d` and `- 5d`

Here, first term `= - 9d`

and common difference `= - 8d + 9d = d`

So, if d is positive, then first term should be negative and

common difference should be positive.

If `d` is negative, then first term should be positive and common

difference should be negative.

Hence, either the first term or the common difference is negative

but not both.

Correct Answer is `=>` (C) Either the first term or the common difference is negative but not both

Q 2307134988

Read the following information carefully and then answer the given questions.

The sum of first `10` terms and `20` terms of an AP are `120`
and `440`, respectively.
What is its first term?
NDA Paper 1 2012

(A)

`2`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

Let the first term of an `AP` be a and

common difference be `d`.

Given, `S_(10) = 120` and `S_(20) = 440`

` ∵ S_n = n/2 [2a + (n -1)d] => S_(10) = (10)/2 [ 2a + (10 -1)d]`

`=> 120 = 5(2a + 9d) => 2a + 9d = 24 `... (i)

and `S_(20) = (20)/2 [2a + (20 - 1)d]`

`=> 440 = 10(2a + 19d) => 2a + 19d = 44` ... (ii)

On subtracting Eq. (i) from Eq. (ii), we get

`10d = 20 => d = 2`

On putting the value of d in Eq. (i), we get

` 2a + 9(2) = 24 => 2a + 18 = 24`

` => 2a = 6`

`:. a = 3`

Correct Answer is `=>` (B) `3`

Q 2347345283

What is the `n^(th)` term of the following sequence?
`1, 5, 9, 13, 17, ...`
NDA Paper 1 2012

(A)

`2n - 1`

(B)

`2n + 1`

(C)

`4n - 3`

(D)

None of these

Solution:

Given, sequence is `< S_n > = 1, 5, 9,13, 17,..`

Since, the common difference is `4` in each consecutive terms,

which forms an `AP`.

Let `a = ` First term `= 1, d =` Common difference `= 4`

Then, `T_n =a+ (n - 1)d` (nth term)

` = 1 + (n - 1)4 = 1 + 4n - 4 = 4n - 3`

Correct Answer is `=>` (C) `4n - 3`

Q 2327756681

The `59`th term of an AP is `449` and the `449`th
term is `59`. Which term is equal to `0` (zero)?
NDA Paper 1 2010

(A)

`501`st term

(B)

`502`nd term

(C)

`508`th term

(D)

`509`th term

Solution:

Let `a` and `d` be the first term and common difference of

the AP.

`:. a + 58 d = 449` ........(i)

and `a + 448 d = 59` .........(ii)

On solving Eqs. (i) and (ii), we get

`a = 507` and `d = - 1`

Now, assume that `n`th term will be zero.

`:. 0 = 507 + (n - 1 ) (-1)`

`=> 507 = n - 1 => n = 508`

Correct Answer is `=>` (C) `508`th term

Q 2337667582

If the sum of 'n' terms of an arithmetic progression
is `n^2 - 2n`, then what is the `n`th term?
NDA Paper 1 2008

(A)

`3n - n^2`

(B)

`2n - 3`

(C)

`2n + 3`

(D)

`2n - 5`

Solution:

Given,

`S_n = n^2 - 2n`

`:. a_n = S_n - S_(n-1)`

`= n^2 - 2n- [(n- 1)^2 - 2(n - 1)]`

`= n^2 - 2n - [n^2 + 1 - 2n - 2n + 2]`

`= 2n - 3`

Correct Answer is `=>` (B) `2n - 3`

Q 2337567482

If the number of terms of an `AP` is `(2n + 1)`, then
what is the ratio of the sum of the odd terms to
the sum of even terms?
NDA Paper 1 2008

(A)

`n/(n + 1)`

(B)

`n^2/(n + 1)`

(C)

`(n + 1)/ n`

(D)

`(n + 1)/( 2n)`

Solution:

Let the `AP` be

`a, a+ d, a+ 2d, ... ,a+ (2n - 1)d, a+ 2nd`

Series of odd terms

`a + d, a+ 3d, ... , a + (2n - 1)d`, has `n` terms.

`:.` Sum `= n/2 [(a+ d)+ {a+ (2n -1)d}]`

`= n/2 [2a +2nd] = n[a + nd]`

Series of even terms

`a, a+ 2d, a+ 4d, ... , a + 2nd,` has `(n + 1)` terms.

`:.` Sum `= (n + 1)/2 [a + (a + 2nd)]`

` = (n + 1)/2 (2a +2nd)`

`= (n+ 1) (a + nd)`

`:.` Requierd ratio `= (n + 1)/n`

Correct Answer is `=>` (C) `(n + 1)/ n`

Q 2307856788

Let `a, b` and `c` be in `AP`. Consider the following
statements.
I . `1/(ab) , 1/(ca)` and `1/(bc)` are in `AP`.
II. `1/(sqrt(b) + sqrt(c) ) , 1/(sqrt(c) + sqrt(a) ) ` and `1/(sqrt(a) + sqrt(b) )` are in `AP`.

Which of the statements given above is/are correct?
NDA Paper 1 2010

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Let `1/(ab), 1/(ca)` and `1/(bc)` be in `AP`.

`=> 1/(ca) - 1/(ab) = 1/(bc) - 1/(ca) => ( b-c)/(abc) = (a-b)/(abc)`

`=> b - c = a - b => 2b = a + c`

So, `a, b` and `c` are in AP.

Now, `1/(sqrt(b) + sqrt(c) ) , 1/(sqrt(c) + sqrt(a) ) ` and `1/(sqrt(a) + sqrt(b) )` are in AP.

` :. 2/(sqrt(c) + sqrt(a) ) = 1/(sqrt(b) + sqrt(c) ) + 1/ (sqrt(a) + sqrt(b) )`

`=> 2( sqrt(b) + sqrt(c)) ( sqrt(a) + sqrt(b)) = ( sqrt(c) + sqrt(a)) ( sqrt(a) + 2 sqrt(b) + sqrt(c))`

`=>2 ( sqrt(ab) + b + sqrt(ac) + sqrt(bc) ) = sqrt(ac) + 2 sqrt(bc) + c + a`

` + 2 sqrt(ab) + sqrt(ac)`

`=>2 sqrt(ab) + 2b + 2 sqrt(ac) + 2 sqrt(bc) = 2 sqrt(ac) + 2 sqrt(bc)`

` + 2 sqrt(ab) + c +a`

` => 2b =a+ c`

So, `a, b` and `c` are in AP.

Hence, both the statements are correct.

Correct Answer is `=>` (C) Both I and II

Q 2337845782

If `a, b, c, d, e` and `f` are in AP, then `(e- c)` is equal to
which one of the following?
NDA Paper 1 2011

(A)

`2(c -a)`

(B)

`2(d- c)`

(C)

`2(f- d)`

(D)

`(d -c)`

Solution:

Given, `a, b, c, d, e` and f are in AP.

`b - a = c - b => b - c = a - b` ... (i)

and `e - d =d - c` ... (ii)

On adding both the equations, we get

`(e -c)+ (b- d)= (a- c)+ (d -b)`

`=> (e-c)=(d-c)+(a+d)-2b` ... (iii)

(`∵` a, b and c are in AP)

`=> 2b =a+ c`

From Eq. (iii),

`=> (e - c) = (d - c)+ (a+ d) - (a+ c)`

`= (d - c) + (d - c)`

`=> (e - c) = 2(d - c)`

Correct Answer is `=>` (B) `2(d- c)`

Q 2337267182

Which term of the sequence `20, 19 1/4 , 18 1/2 , 17 3/4` ,
... , is the first negative term?
NDA Paper 1 2009

(A)

27th

(B)

28th

(C)

29th

(D)

No such term exists

Solution:

Given, series can be rewritten as

` 20 , (77)/4 , ( 37)/2 , (71)/4 , .....`

This is an `AP` series.

Here, `a = 20, d = - 3/4`

`:. T_n = a + (n - 1)d`

`= 20 + (n - 1) (- 3/4) = (83)/4 - 3/4 n`

For first negative term, `T_n < 0 => (83)/4 - 3/4 n < 0`

` => 83 < 3n => n > (83)/3`

So, `n` should be `28`.

Hence, `28`th term is the first negative term.

Correct Answer is `=>` (B) 28th

Q 2347056883

If `p` times the `p` th term of an AP is `q` times the `q` th
term, then what is the `(p + q)` th term equal to?
NDA Paper 1 2010

(A)

`p + q`

(B)

`pq`

(C)

`1`

(D)

`0`

Solution:

Let a and d be the first term and common difference of

an AP.

According to the question,

`p·T_p = q·T_q => p[a+(p-1)d] = q[a+(q - 1)d]`

`=> pa + (p^2 - p)d = qa + (q^2 - q )d`

`=> (p- q )a = (q^2 - p^2 + p- q )d`

`=> (p-q)a = (p-a) (-p-q+1)d => a = -(p+q-1)d`

Now, `T_(p +q) = a + (p + q - 1)d`

`= - (p + q - 1)d + (p + q - 1)d = 0`

Correct Answer is `=>` (D) `0`

Problems on AM

`AM = (a+b)/2`

Q 2357067884

The arithmetic mean of `4` numbers is `15`. The
arithmetic mean of another `6` numbers is `12`.
What is the arithmetic mean of the combined `10`
numbers?
NDA Paper 1 2008

(A)

`12.2`

(B)

`12.8`

(C)

`13.2`

(D)

`13.8`

Solution:

Arithmetic mean of the combined `10` numbers

`= (4 xx 15 + 6 xx 12)/(10) quad (∵ bar x_(12) = (bar x_1 n_1 + bar x_2 n_2)/(n_1 + n_2))`

where, ` = 4, bar x_1 = 15` and `n_2 = 6, bar x_2 = 12`

` = (60 + 72)/(10) = 13.2`

Correct Answer is `=>` (C) `13.2`

Q 1147645583

`‘n’` arithmetic means are inserted between the numbers `7` and `49`. If the sum of these means be `364`, then find the sum of their square

(A)

`16250`

(B)

`11830`

(C)

`10360`

(D)

`14280`

Solution:

`A_1+A_2+......+A_n = n ((7+49)/2)= 364 `

`⇒ n = 3.`

Hence, `49 = 7 + 144d `

`⇒d = 3`

`13` A.M.'s are given by `7 + 3k, k = 1, 2, ........,13`

∴ Required sum = `sum_(k=1)^13(7+3k)^2=9sum_(k=1)^13 k^2+42sum_(k=1)^13 k+49 sum_(k=1)^13(1)`

`= 11830.`

Correct Answer is `=>` (B) `11830`

Q 1417223189

Given that n arithmetic means are inserted between two
sets of numbers `a, 2b` and `2a, b`, where `a, b in R.` Suppose
further that mth mean between these two sets of numbers
is same, then the ratio, `a : b` equals

(A)

`n - m + 1: m`

(B)

`n - m + 1 : n`

(C)

`m : n - m + 1`

(D)

`n : n - m + 1`

Solution:

` a, A_1, A_2, ......., A_n, 2b`

`2b = (n+2)`th term = `a+(n+2-1)d`

`d = (2b-a) /(n+1)d`

` A_m = a + m (( 2b+a )/ (n+1)) `

Again `2a, B_1, B_2, ........ , B_n , b`

`b = (n + 2)`th term =` 2a + (n + 2 -1) D`

`D = (b- 2a)/(n+1) `

`:. B =2a + mD =2a + m ((b- 2a)/(n+1 ))`

`=> A_m = B_m`

`=> a+ m ((2b - a)/ (n+1 )) = 2a + m((b- 2a)/(n+1 ))`

` => m (a+ b) = a (n + 1)`

or `a (n - m + 1) = bm`

`:. a : b = m : n - m + 1`

Correct Answer is `=>` (C) `m : n - m + 1`

Q 2357245184

What is the arithmetic mean of first `16` natural
numbers with weights being the number itself?
NDA Paper 1 2012

(A)

`17//2`

(B)

`33//2`

(C)

`11`

(D)

`187//2`

Solution:

We know that, the arithmetic mean of `n` natural

numbers with weights being the number itself

` = (sum n^2)/(sum n) = (( n(n + 1) (2n + 1))/6)/((n ( n + 1))/2)`

`= (n (n + 1) (2n + 1))/6 xx 2/(n(n + 1)) = (2n + 1)/3`

For `16` natural numbers, put `n = 16`

`= (2 xx 16 + 1)/3 = (33)/3 = 11`

Correct Answer is `=>` (C) `11`

Problem On GP

`T_n = a . r^(n-1)`

`S=(a(1-r^n))/(1-r)` ,where `r ne 1`

`S_(oo)=a/(1-r)`

Q 2723191041

The sum of the first n terms of the series `1/2 + 3/4 + 7/8 + (15)/(16) + ...` is equal to
NDA Paper 1 2017

(A)

`2^n - n - 1`

(B)

`1 - 2^(-n)`

(C)

`2^(-n) + n - 1`

(D)

`2^n - 1`

Solution:

`1/2 +3/4 + 7/8 +15/16+............`

`=(1-1/2) +(1-1/4)+ (1- 1/8) +.............n` terms

`=(1+1+.........1 ` n terms ) - (`1/2 +1/4 +1/8 +.............n` terms)

`=n - {(1/2 (1- (1/2)^n)/((1-1/2))}`

`= 2^(-n) +n-1`

Correct Answer is `=>` (C) `2^(-n) + n - 1`

Q 2116112970

Given that `log_(x) y, log_(2) x, log_(y) z` are in `GP, xyz = 64` and ` x^(3) , y^(3) ,z^(3) `
are in `AP`.

Which one of the following is correct?
x, y and z are
NDA Paper 1 2016

(A)

in AP only

(B)

in GP only

(C)

in both AP and GP

(D)

Neither in AP nor in GP

Solution:

Given, `log_(x) y, log_(z) x, log_(y) z` are in G P.

`=> (log_( z) x)^(2) = log_(x) y xx log _(y) z`

`=> log_(x) z => 1/(log _(z) x)`

`=> (log_(z)x)^(3) = 1 => log _(2) x = 1`

` => x= z`

Now, `x^(3), y^( 3), z^(3)` are inAP.

`:. 2y^(3) = x^(3) + z^(3)`

`=> 2y^(3) = z^(3) + z^(3)`

`=> y^(3) = z^( 3)`

`:. y= z`

`:. x = y = z`

Also, `xyz = 64 => xyz =, 4^(3)`

`=> x= y = z = 4`

Correct Answer is `=>` (C) in both AP and GP

Q 2701867728

What is the greater value of the positive integers n satisfying the condition `1 + 1/2 + 1/4 + 1/8 + ... + 1/ ( 2^( n -1) ) < 2 - 1/1000 `?
NDA Paper 1 2016

(A)

(B)

(C)

(D)

Solution:

`1* (1/2)^0 + (1/2)^1 + (1/2)^2 + (1/2)^3 + ....... (1/2)^(n-1) < 2 - 1/1000`

`(1 * (1- (1/2)^n ) )/(1-1/2) < 2- 1/1000`

`2-2 (1/2)^n < 2 - 1/1000`

`(1/2)^n > 1/(2 xx 1000)= 1/(2 xx 2^9 xx 1 .95)`

` (1/2)^n > (1/2)^10 xx 1/(1.95)`

on comparing ` n > 10`

`n =11`

Correct Answer is `=>` (D) 11

Q 2117580489

If `m` is the geometric mean of
`(y/z)^(log(yz)) , (z/x)^(log(zx)) ` and `(x/y)^(log(xy))`
then what is the value of `m`?
NDA Paper 1 2016

(A)

`1`

(B)

`3`

(C)

`6`

(D)

`9`

Solution:

Here, `m = [ (y/z)^(log(yz)) xx (z/x)^(log(zx)) xx (x/y)^(log(xy)) ]^(1//3)`

`:. m^3 = x^( log (xy) - log (zx)) xx y^(log (yx) - log (xy) ) xx z^(log (zx) - log (yz))`

` => m^3 = x^( log (y/z)) xx y^( log (z/x)) xx z^( log (x/y))`

Taking log on both sides, we get

`3 log m = log ( y/z) log x + log (z/x) log y + log (x/y) log z`

`=> 3 log m = log y log x - log z log x + log z log y`

`- log x log y + log x log z - log y log z`

`=> 3 log m = 0`

`=> log m = 0 => m = e^0 = m = 1`

Correct Answer is `=>` (A) `1`

Q 2201312228

If `p, q, r` are in one geometric progression and `a, b, c` are in
another geometric progression, then `ap, bq, cr` are in
NDA Paper 1 2015

(A)

`AP`

(B)

`GP`

(C)

`HP`

(D)

None of these

Solution:

Clearly, `ap, bq, cr` are in `GP` because on multiplying

corresponding terms of two `GP's` and the resulting

series is also a `GP`.

e.g. `S_1 = 2, sqrt(2), 1, 1/sqrt(2)` , .......

`S_2 = 3, sqrt(3) , 1, 1/sqrt(3)` , .......

`S _3 = 6, sqrt(6), 1/sqrt(6)` , .......

Here `alpha = 3 r =sqrt(6)/6 = 1/sqrt(6)`

Correct Answer is `=>` (B) `GP`

Q 2317578480

If `1, x, y, z` and `16` are in geometric progression,
then what is the value of `x + y + z`?
NDA Paper 1 2007

(A)

`8`

(B)

`12`

(C)

`14`

(D)

`16`

Solution:

Since, `1, x, y, z` and `16` are in geometric progression.

Here, `a = 1, l = 16, n = 5`

`l = ar^(n - 1)`

`16 = 1 · r^4 => r = 2`

`:. x = 1· r = 2, y = 1· r^2 = 4`,

`z = 1·r^3 = 8`

`:. x + y + z =2 + 4 +8 = 14`

Correct Answer is `=>` (C) `14`

Q 2347545483

What is the sum of the series `1 - 1/2 + 1/4 - 1/8 + · · ·` ?
NDA Paper 1 2012

(A)

`1//2`

(B)

`3//2`

(C)

`2`

(D)

`2//3`

Solution:

Given series is `1 - 1/2 + 1/4 - 1/8 + ...`

which form a `GP` with common ratio `(- 1/2)`

`:.` Sum of infinite term of `GP = a/(1 - r) = 1/(1 - ((-1)/2))`

`= 1/ (1 + 1//2) = 2/3`

Correct Answer is `=>` (D) `2//3`

Q 2337767682

If `a, 2a + 2, 3a + 3` are in `GP`, then what is the `4`th
term of that series?
NDA Paper 1 2008

(A)

`- 13.5`

(B)

`13.5`

(C)

`-27`

(D)

`27`

Solution:

` ∵ a. 2a + 2` and `3a + 3` are in GP.

`:. (2a + 2)^2 = a(3a + 3) => 4a^2 + 4 + 8a = 3a^2 + 3a`

`=> a^2 + 5a + 4 = 0 => (a + 4)(a + 1) = 0`

`=> a= -1` or `-4`

Let the `4`th term be `x`.

`:. a/(2a + 2) =(3a+3)/x => x = ((3a+3)(2a+2))/a`

When `a =- 4`, then `x = -13.5`

and when `a = - 1` then `x = 0`

So, the `4`th term is `-13.5`.

Correct Answer is `=>` (A) `- 13.5`

Q 2357045884

Which term of a series `1/4 ,- 1/2 , 1 , ...` is `-128`?
NDA Paper 1 2011

(A)

9th

(B)

10th

(C)

11th

(D)

12th

Solution:

` ∵ a = 1/4 , r = - (- 1//2)/(1//4) = -2`

So, the given series forms a GP.

`:. T_n = ar^(n - 1) => -128 = 1/4 (-2)^(n - 1)` (given)

`=> (-2)^9 = (-2)^(n - 1) => 9 = n - 1`

`=> n = 10`

Correct Answer is `=>` (B) 10th

Q 2307145988

What is the sum of `sqrt(3) + 1/sqrt(3) + 1/(3sqrt3) + .. `?
NDA Paper 1 2011

(A)

`sqrt3/2 `

(B)

`(3 sqrt3)/2`

(C)

`(2 sqrt3)/3`

(D)

`sqrt 3`

Solution:

`sqrt(3) + 1/sqrt(3) + 1/(3sqrt3) + .. quad ( ∵ S_(oo) = a/( 1 - r))`

` = sqrt3/(1 - 1/3) = (3sqrt3)/2`

Correct Answer is `=>` (B) `(3 sqrt3)/2`

Q 2701145928

How many geometric progressions is/are possible containing 27, 8 and 12 as three of its/their terms ?
NDA Paper 1 2016

(A)

one

(B)

two

(C)

four

(D)

Infinitely many

Solution:

Let term of GP

`T_p= 27 = ar^(p-1)`...............(1)

`T_q = 8 = ar ^(q-1)`................(2)

`T_s= 12= ar^(s-1)`......................(3)

From (1) and (2)

`r^(p-q) = 27/8 => r^(p-q) =(3/2)^3`

From (2) and (3)

`r^(s-p) = 8/12 => r^(s-p)= (2/3)^1 =(3/2)^(-1)`

From (3) and (4)

`r^(s-p) = 12/27 => r^(s-p) = 4/9 = (3/2)^(-2)`

so, `r= 3/2 , p-q=3, q-s=-1, s-p=-2`

As there can be infinite natural number for satisfying these equation.

`:. ` There can be infinite GP

Correct Answer is `=>` (A) one

Q 2242801733

The geometric mean of the observations `x_1 , x_2 , x_3 , .... , x_n`
is `G_1`. The geometric mean of the observations
`y_1,y_2 ,y_3 , ... ,y_n` is `G_2` . The geometric mean of
observations ` x_1/y_1 , x_2/y_2 , x_3/y_3 , .. , x_n /y_n` is

NDA Paper 1 2015

(A)

`G_1G_2`

(B)

ln `(G_1G_2)`

(C)

`G_1/G_2`

(D)

ln `(G_1/G_2)`

Solution:

Geometric mean of `x_1 , x_2, x_3 , ... , x_n` is `G_1`

`=> G_1 = ( x_1 , x_2, ... , x_n)^(1/n)`

Geometric mean of `y_1, y_2 , y_3, .... ,y_n` is `G_2`

`=> G_2 = ( y_1, y_2 , y_3, .... ,y_n)^(1//n)`

`:. GM` of `x_1/y_1, x_2/y_2 , x_3/y_3 , .... , x_n/y_n `

`= ( x_1/y_1, x_2/y_2 , x_3/y_3 , .... , x_n/y_n )^n = ( (x_1 ,x_2 , x_3 ,.., x_n))^(1//n)/ ( (y_1 ,y_2 , y_3 ,.., y_n))^(1//n) = G_1/G_2`

Correct Answer is `=>` (C) `G_1/G_2`

Q 2146123073

Given that `log_(x) y, log_(2) x, log_(y) z` are in `GP, xyz = 64` and ` x^(3) , y^(3) ,z^(3) `
are in `AP`.

Which one of the following is correct?
xy, yz and zx are
NDA Paper 1 2016

(A)

in AP only

(B)

in GP only

(C)

in both AP and GP

(D)

Neither in AP nor in GP

Solution:

Given, `log_(x) y, log_(z) x, log_(y) z` are in G P.

`=> (log_( z) x)^(2) = log_(x) y xx log _(y) z`

`=> log_(x) z => 1/(log _(z) x)`

`=> (log_(z)x)^(3) = 1 => log _(2) x = 1`

` => x= z`

Now, `x^(3), y^( 3), z^(3)` are in AP.

`:. 2y^(3) = x^(3) + z^(3)`

`=> 2y^(3) = z^(3) + z^(3)`

`=> y^(3) = z^( 3)`

`:. y= z`

`:. x = y = z`

Also, `xyz = 64 => xyz =, 4^(3)`

`=> x= y = z = 4`

`xy, yz` and `zx` are in both AP and GP.

Correct Answer is `=>` (C) in both AP and GP

Q 2327278181

The product of first nine terms of a `GP` is, in
general, equal to which one of the following?
NDA Paper 1 2008

(A)

The 9th power of the 4th term

(B)

The 4th power of the 9th term

(C)

The 5th power of the 9th term

(D)

The 9th power of the 5th term

Solution:

First nine terms of a GP are `a, ar, ar^2, ... , ar^8`

`:. P = a · ar · ar^2 ..... ar^8`

`= a^9 . r^(1 + 2+ ..... + 8 ) = a^9 . r^(8/2 (1 + 8)) `

`= a^9 . r^((8.9)/ 2) = a^9r^(36) = (ar^4)^9 = (T_5)^9`

`= 9`th power of the `5`th term

Correct Answer is `=>` (D) The 9th power of the 5th term

Q 2201312228

If `p, q, r` are in one geometric progression and `a, b, c` are in
another geometric progression, then `ap, bq, cr` are in
NDA Paper 1 2015

(A)

`AP`

(B)

`GP`

(C)

`HP`

(D)

None of these

Correct Answer is `=>` (B) `GP`

Q 2357167984

What is sum to the `100` terms of the series
`9 + 99 + 999 + ...` ?
NDA Paper 1 2008

(A)

`(10)/9 (10^(100) -1) - 100`

(B)

`(10)/9 (10^(99) -1) - 100`

(C)

`100 (100^(10) -1)`

(D)

` 9/(100) ( 10^(100) - 1)`

Solution:

Let `S = 9 + 99 + 999 + ...`

`= (10^1 - 1) + (10^2 - 1) + (10^3 - 1) + ...`

`= (10 + 10^2 + 10^3 + ... )`

`- (1 + 1 + 1 + ... + 100` times)

` = (10(10^(100) - 1))/( 10 - 1) - 100`

` = (10)/9 (10^(100) -1) - 100`

Correct Answer is `=>` (A) `(10)/9 (10^(100) -1) - 100`

Q 2347367283

In a geometric progression with first term `a` and
common ratio `r`, what is the arithmetic mean of
first `5` terms?
NDA Paper 1 2009

(A)

`a+ 2r`

(B)

`ar^2`

(C)

`a(r^5 - 1)// (r- 1) 5`

(D)

`a(r^5 - 1) // [5(r- 1)]`

Solution:

Let first `5` terms of a geometric progression are `a, ar`,

`ar^2 , ar^3` and `ar^4`

`:.` Mean `= (a + ar + ar^2 + ar^3 + ar^4)/5`

` = (a (1 + r + r^2 + r^3 + r^4))/5`

` = (a (( r^5 - 1)/(r - 1)))/5`

`= (a(r^5 - 1))/(5 (r- 1))`

Correct Answer is `=>` (C) `a(r^5 - 1)// (r- 1) 5`

Q 2337445382

If the sequence `{S_n}` is a geometric progression
and `S_2S_(11) = S_p S_8`, then what is the value of `p`?
NDA Paper 1 2012

(A)

(B)

(C)

(D)

Cannot be determined

Solution:

We know that, in a `GP` the product of two terms

equidistant from the beginning and end is a constant and is equal

to the product of first term and last term i.e., if

`a_1 ,a_2 ,a_3, ... ,a_((n - 2)) , (a_(n- 1)) a_n` are in GP, then

`a_1 a_n = a_2 a_(n - 1) = a_3 a_(n- 2) = ...`

Given that, `S_2S_(11) = S_pS_8 `

` => (p + 8) = (2 + 11)`

`=> p = 13 - 8 = 5`

Correct Answer is `=>` (C) 5

Q 2367634585

What is `0.9 + 0.09 + 0.009 + ...` equal to?
NDA Paper 1 2013

(A)

`1`

(B)

`1.01`

(C)

`1.001`

(D)

`1.1`

Solution:

Series `= 0.9 + 0.09 + 0.009 + ...`

`= 9{0.1 + 0.01 + 0.001 + ...}`

`= 9 { 1/(10) + 1/(100) + 1/(1000) + ...}`

`= 9 {10^(-1) + 10^(-2) + 10^(-3) + ... }`

`= 9/(10) {1 + (1/(10))^1 + ( 1/(10))^2 + ... }`

[which form an infinite `GP` with common ratio `( 1/(10))`]

`= 9/(10) xx 1/(1 - 1/(10)) = 9/(10) xx (10)/9 = 1`

Correct Answer is `=>` (A) `1`

Q 2367134985

What is the sum of first `8` terms of the series
` 1 - 1/2 + 1/4 - 1/8 + ...` ?
NDA Paper 1 2012

(A)

`(89)/(128)`

(B)

`(57)/(384)`

(C)

`(85)/(128)`

(D)

None of these

Solution:

Given series is

`1 - 1/2 + 1/4 - 1/8 + ...`

Since, it is a geometric progression.

Here, first term `a = 1` and common ratio `r = - 1/2 < 1`.

`:.` The sum of first `8` terms of the series

i.e., `S_8 = ( a(1- r^8))/(1 - r)`

[by formula, `S_n = ( a(1 - r^n))/(1 - r)`, when `r < 1`]

`= ( 1 [1 - (- 1/2)^8 ])/(1 - (- 1/2)) = ( 1- 1/(256))/(1 + 1/2) = (255//256)/(3//2)`

` = (255)/(256) xx 2/3 = (85)/(128)`

Correct Answer is `=>` (C) `(85)/(128)`

Q 2317167089

If `x, 2x + 2` and `3x + 3` are the first three terms of a
GP, then what is its fourth term?
NDA Paper 1 2009

(A)

`- 27//2`

(B)

`27//2`

(C)

`- 33//2`

(D)

`33//2`

Solution:

Since, `(2x + 2)^2 = x(3x + 3) => x^2 + 5x + 4 = 0`

`=> x = - 1` and `-4`

Now, first term `a = x`

Second term, `ar = 2 x + r = (2(x + 1))/x`

`:.` Fourth term `= ar^3 = x ((x +1)/x)^3`

Put ` x = -4`

` T_4 = -4 ( (2(-4 + 1))/(-4))^3 = -4 xx ( 3/2)^3 = - (27)/2`

Correct Answer is `=>` (A) `- 27//2`

Q 2307445388

If p, q and r are in AP as well as GP, then which
one of the following is correct?
NDA Paper 1 2012

(A)

`p = q != r`

(B)

`p != q != r`

(C)

`p != q = r`

(D)

`p = q = r`

Solution:

Given that, `p, q` and `r` are in AP.

`:. 2q = p + r` ........(i)

As well as are in GP.

`:. q^2 = pr` ........(ii)

From Eqs. (i) and (ii),

`p + r = 2 sqrt(pr)`

` => (sqrt p )^2 - 2 sqrt p . sqrt r + (sqrt r)^2 = 0`

`=> ( sqrt p - sqrt r )^2 = 0`

` => sqrt p - sqrt r => p = r` ........(iii)

From Eq. (ii),

`q^2 = r·r = r^2 = q = r` .........(iv)

From Eqs. (iii) and (iv),

` p = q = r`

Correct Answer is `=>` (D) `p = q = r`

Q 2367645585

If `n !,3 xx (n !)` and `(n + 1)!` are in GP, then the value
of `n` will be
NDA Paper 1 2011

(A)

`3`

(B)

`4`

(C)

`8`

(D)

`10`

Solution:

Given that, `n ! , 3 xx (n !)` and `(n + 1) !` are in GP.

Then, `{3 xx (n !)}^2 = (n !) xx (n + 1)!`

`=> (3)^2 (n !) = (n + 1) !`

`=> 9(n!) = (n + 1)·(n!)`

`=> 9 = n + 1 => n = 8`

Correct Answer is `=>` (C) `8`

Q 2387745687

If the `10`th term of a GP is `9` and `4`th term is `4`,
then what is its `7`th term?
NDA Paper 1 2011

(A)

`6`

(B)

`14`

(C)

`27//14`

(D)

`56//15`

Solution:

Given, `10`th term of `GP = 9`

`T_(10) = ar^(10- 1) = 9 => ar^9 = 9` ... (i)

(where, `a` = first term and `r` = common difference)

and `4`th term of `GP = 4`

`T_4 = ar^(4- 1) = ar^3 = 4` ... (ii)

On dividing Eq. (i) by Eq. (ii), we get

`(ar^9)/(ar^3) = 9/4 => r^6 = 9/4` ......(iii)

Now, `7`th term of GP is

`T_7 = ar^6 = (9a)/4` ... (iv)

On multiplying Eqs. (i) and (ii), we get

`(ar^9) (ar^3) = 9· 4 => a^2r^(12) = 36`

`=> a^2(r^6)^2 = 36`

`=> a^2 ( 9/4 )^2 = 36 => a^2 = (36 xx 16)/(81)`

` => a = ( 6 xx 4)/9 => a = 8/3`

From Eq. (iv),

` T_7 = 9/4 xx 8/3 = 2 xx 3 = 6`

Correct Answer is `=>` (A) `6`

Q 2317845789

In a `GP` of positive terms, any term is equal to
one-third of the sum of next two terms. What is
the common ratio of the `GP`?
NDA Paper 1 2011

(A)

`(sqrt(13) + 1)/2`

(B)

`(sqrt(13) - 1)/2`

(C)

`(sqrt(13) + 1)/3`

(D)

`sqrt(13)`

Solution:

If `a, ar, ar^2, ...` are in GP, then

According to the question,

`a = 1/3 (ar + ar^2) => 3 = r + r^2`

`=> r^2 + r - 3 = 0 => r = ( -1 ± sqrt(1 + 4 xx 3))/2`

`=> r = ( -1 pm sqrt(13))/2 = (sqrt(13) - 1)/2 quad (∵ r > 0)`

Correct Answer is `=>` (B) `(sqrt(13) - 1)/2`

Q 2347356283

A square is drawn by joining mid-points of the
sides of a square. Another square is drawn inside
the second square in the same way and the process
is continued indefinitely. If the side of the first
square is `16 cm`, then what is the sum of the areas
of all the squares?
NDA Paper 1 2010

(A)

`256` sq cm

(B)

`512` sq cm

(C)

`1024` sq cm

(D)

`512//3` sq cm

Solution:

Required sum

`= (16)^2 + 1/2 (16)^2 + 1/4 (16)^2 + ... oo`

`= (16)^2 { 1 + 1/2 + 1/4 + ... oo}`

`= (16)^2 { 1/( 1- 1/2) } = 256 xx 2 = 512` sq cm

Correct Answer is `=>` (B) `512` sq cm

Q 2337456382

The sum of an infinite geometric progression is `6`.
If the sum of the first two terms is `9//2`, then what
is the first term?
NDA Paper 1 2010

(A)

`1`

(B)

`5//2`

(C)

`3` or `3//2`

(D)

`9` or `3`

Solution:

`∵ a/(1 - r) = 6 => a = 6(1 - r)` .......(i)

and `a + ar = 9/2` (given)

` => 6(1 - r) + 6r(1 - r) = 9/2`[fromEq.(i)]

`=> 12 -12r + 12r -12r^2 = 9`

`=> r^2 = 3/(12) = 1/4 => r = 1/2 ` or `r = 1/2 => a = 3` or `9`

Correct Answer is `=>` (D) `9` or `3`

Problems on GM

`GM = sqrt(ab)`

Q 2242801733

(A)

`G_1G_2`

(B)

ln `(G_1G_2)`

(C)

`G_1/G_2`

(D)

ln `(G_1/G_2)`

Correct Answer is `=>` (C) `G_1/G_2`

Q 2337056882

The geometric mean of three numbers was
computed as 6. It was subsequently found that in
this computation, a number 8 was wrongly read as
12. What is the correct geometric mean?
NDA Paper 1 2010

(A)

`4`

(B)

`root(3)(5)`

(C)

`2 root(3)(18)`

(D)

None of these

Solution:

When we take 12 wrongly in place of 8, then geometric
mean= 6

`(x_1 *x_2 * 12)^(1/3) = 6`

`=> x_1 * x_2 *12 = 216`

`=> x_1 * x_2 = 18` ...........(i)

Now, we take the right observation 8 in place of 12, then the
geometric mean ` = (x_1 * x_2 * 8)^(1/3)`

` = (18 *8)^(1/3) = 2 root(3)(18)`

Correct Answer is `=>` (C) `2 root(3)(18)`

Q 2367845785

If the arithmetic and geometric means of two
numbers are `10, 8` respectively, then one number
exceeds the other number by
NDA Paper 1 2011

(A)

`8`

(B)

`10`

(C)

`12`

(D)

`16`

Solution:

Arithmetic mean of `a` and `b` is `(a + b)/2`.

`=> (a+ b)/2 = 10`(given)

`=> a + b = 20` .....(i)

Geometric mean of `a` and `b` is `sqrt(ab)`.

`=> sqrt(ab) = 8`(given)

`=> ab = 64` ..(ii)

Now, we have `(a- b)^2 = (a+ b)^2 - 4ab`

`= (20)^2 - 4(64)`

`= 400- 256= 144`

`=> a - b = 12` ......(iii)

From Eqs. (i) and (iii),

`a= 16` and `b = 4`

So, one number exceeds the other by `12`.

Correct Answer is `=>` (C) `12`

Q 2327845781

What is the geometric mean of `1 0, 40` and `60?`
NDA Paper 1 2011

(A)

`10 root(3)(3)`

(B)

`20 root(3)(3)`

(C)

`40 root(3)(3)`

(D)

`70 root(3)(3)`

Solution:

The geometric mean of the given observation

`(10*40*60)^(-1/3) = (24xx1000)^(1/3)`

` = (3)^(1/3) (8xx1000)^(1/3) = (3)^(1/3)(2xx10) = 20 root(3)(3)`

Correct Answer is `=>` (B) `20 root(3)(3)`

Q 2357145084

What is the geometric mean of the sequence
`1, 2, 4, 8, ... ,2^n`?
NDA Paper 1 2012

(A)

`2^(n//2)`

(B)

`2^((n+ 1)//2)`

(C)

`2^(n+ 1) - 1`

(D)

`2^(n - 1)`

Solution:

The geometric mean of `1, 2, 4, 8, ... , 2^n`

`GM = (1·2·4·8· ... ·2^n)^(1/(n +1))`

`= (2 ·2^2 ·2^3 · ... · 2^n )^(1/(n +1))`

`= (2^(1 + 2+ 3+ ... + n ) )^(1/(n +1)) = (2 ^(sum n) )^(1/(n +1))`

`= 2^((n (n+1))/2 xx 1/(n + 1)) = 2^(n//2)`

Correct Answer is `=>` (A) `2^(n//2)`

Q 2554745654

If `a+ 2b + 3c = 12, (a, b, c in R^+)`, then `ab^2c^3` is
UPSEE 2008

(A)

` ge 2^3`

(B)

` ge 2^6`

(C)

` le 2^6`

(D)

None of these

Solution:

Given that, `a+ 2b + 3c = 12` ... (i)

and `a, b, c` are positive real numbers.

Now, `AM ge GM`

`=> (a+b +b + c+ c+c)/6 ge root 6 (ab^2 c^3)`

`=> (a+ 2b + 3c)/6 ge root 6 (ab^2 c^3)`

`=> ab^2 x^3 le 2^6`

Correct Answer is `=>` (C) ` le 2^6`

Q 1561434325

Let `alpha, beta, gamma` and `delta` are four positive real number such that their product is unity, then the least value of

`(1 +alpha) (1+beta) (1+gamma)(1 + delta)` is :
BITSAT 2005

(A)

`6`

(B)

`16`

(C)

`0`

(D)

`32`

Solution:

Given that, it is given

`alpha beta gamma delta =1` ....................(i)

As, we know `AM >= GM`

`=> (1+alpha)/2 >= sqrtalpha`

`=> 1+ alpha >= 2sqrtalpha`......................(ii)

Similarly, `1+beta >= 2sqrt beta`...................(iii)

`1+gamma >= 2sqrtgamma` ........................(iv)

and `1 + delta >= 2sqrtdelta`.............(v)

Multiplying Eqs. (ii), (iii), (iv) and (v), we get

`(1 +alpha)(1 +beta)(1 +gamma)(1 + delta) >= 16 sqrt(alpha beta gamma delta)`

`=> (1 +alpha) (1 +beta) (1 +gamma)(1 +delta) = 16`

Correct Answer is `=>` (B) `16`

Q 2805180068

If one AM `'A'` and two GM `p` and `q` be inserted between any two numbers. then the value of `p^3 + q^3` is

(A)

`(2pq)/A`

(B)

`2Apq`

(C)

`2Ap^2q^2`

(D)

None of these

Solution:

A is the arithmetic mean of `a` and `b`.

`therefore A = (a+b)/2`

`p` and `q` are two geometric means between `a` and `b`

`therefore p = a ( b/a)^(1/3) = a^(2/3) * b^(1/3)` and `q = a ( b/a)^(2/3) = a^(1/3) b^(2/3)`

`=> p^3+q^3 = a^2 b +ab^2 = ab (a+b)`

` = 2A ab = 2A pq`

Correct Answer is `=>` (B) `2Apq`

Q 2845180063

Let `a, b, c` be three positive real numbers such that their product is unity, then the least value of `(1 + a)(1 + b)(1 +c)` is

(A)

`16`

(B)

`8`

(C)

`0`

(D)

`3`

Solution:

Since, `abc = 1`

As, we know `AM ge GM`

`=> (1+a)/2 ge sqrta`

`=> 1+a ge 2 sqrta`..............(i)

`=> 1+b ge 2 sqrtb` ...............(ii)
and `1+c ge 2 sqrtc` ..............(iii)

On multiplying Eqs. (i), (ii) and (iii),

we get `(1+a) (1+b) (1+c) ge 8 sqrt(abc)`

`therefore` Least value of `(1+a) (1+b)(1+c) = 8`

Correct Answer is `=>` (B) `8`

Problem On AGP

`T_n = [a + (n - 1) d] r^(n-1)`

` S_(oo)=a/(1-r)+(dr)/((1-r)^2)`

Q 2387767687

What is the `15`th term of the series `3, 7, 3, 21`,
`31,43, ...` ?
NDA Paper 1 2008

(A)

`205`

(B)

`225`

(C)

`238`

(D)

`241`

Solution:

Let `S = 3 + 7 + 13 + 21 + 31 + ... + a_n`

`-S = 3+7+13+21+ ... +a_(n- 1) +a_n`
_____________________________

` 0 = (3 + 4 + 6 + 8 + 10 + 12 + ... + n` terms) `- a_n`

`=> a+n = 3 + { 4 + 6 + 8 + ... + (n - 1)` terms}

`=3+ (n - 1)/2 {8+(n - 1 - 1)2}`

`= 3 + ( n - 1)/2 . 2 { 4 + n - 2}`

`= 3 + (n - 1)(n + 2)`

`:. 15`th term `= a_(15) = 3 + (15 -1)(15 + 2)`

` = 3 + 14·17`

`= 3 + 238 = 241`

Correct Answer is `=>` (D) `241`

Q 1645712663

The sum of n terms of the series `1/(1.3) + 1/(3.5) + 1/(5.7) + ...`
BITSAT 2016

(A)

`1/(2n + 1)`

(B)

`(2n)/(2n + 1)`

(C)

`n/(2n + 1)`

(D)

`(2n)/(n + 1)`

Solution:

Let `T_r` be the `r` th term of the given series. Then,

`T_r = 1/((2r - 1)(2r+1)) , r = 1,2,3......n`

` => T_r = 1/2 (1/(2r-1) - 1/(2r+1))`

Required sum `S` is given by

`S = Sigma_(r = 1)^n T_r`

` => S = 1/2[(1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + ..........+ (1/(2n-1) - 1/(2n+1))]`

` => S = 1/2 [1-1/(2n+1)] = n/(2n+1)`

Correct Answer is `=>` (C) `n/(2n + 1)`

Q 1771523426

If `(10)^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+\cdots +10(11)^{9}=k(10)^{9}`, then k is equal to
JEE Mains 2014

(A)

` \frac{121}{10}`

(B)

` \frac{441}{100}`

(C)

`100`

(D)

` 110`

Solution:

`S=10^{9}+2.11^{1}.10^{8}+\cdots +10.11^{9}`

` \frac{11}{10}.S=11^{1}.10^{8}+\cdots 9.11^{9}+11^{10}`

`\Rightarrow -\frac{1}{10}S=10^{9}+11^{1}.10^{8}+11^{2}.10^{7}+\cdots +11^{9}-11^{10}`

`\Rightarrow -\frac{1}{10}S=10^{9} ( \frac{( \frac{11}{10})^{10}-1}{ \frac{11}{10}-1} )-11^{10}\Rightarrow -\frac{1}{10}S=11^{10}-10^{10}-11^{10}`

`S=10^{11}`

`S=100.10^{9}`

`\Rightarrow k=100`

Correct Answer is `=>` (C) `100`

Q 2327167081

Natural numbers are divided into groups as
`(1), (2, 3), (4, 5, 6), (7,8, 9, 10)` and so on. What
is the sum of the numbers in the `11`th group?
NDA Paper 1 2009

(A)

`605`

(B)

`615`

(C)

`671`

(D)

`693`

Solution:

Given that, the group of natural numbers

`(1), (2, 3), (4, 5, 6), (7, 8, 9, 10), ...`

whose number of terms in each group are

`1,2, 4, 7, ...`

Let `S = 1 + 2 + 4 + 7 + ... + t_n`

and `S = quad 1 + 2 + 4 + ... + t_(n- 1) + t_n`
________________________________
`0 = 1 + 1 + 2 +3 + ... + t_n - t_(n-1) + t_n`

`t_n = 1 + (1 + 2 + 3 + ... + (n - 1) `terms)

`t_n = 1 + ((n - 1))/2 {2 + (n - 2)}`

`t_n = 1 + (n(n - 1))/2`

Therefore, the first term in `11`th group of natural numbers is

`t_(11) = 1 + (11.10)/2 = 56`

So, the 11th group is `(56, , 57, 58, 59, 60, 61, 62, 63, 64, 65, 66)`.

Sum of the terms of 11th group

`S = (11)/2 {2 xx 56 + (11- 1)1}`

`= (11)/2 (112 + 10) = 11 xx 61`

` = 671`

Correct Answer is `=>` (C) `671`

Q 2337745682

What is the `10`th common term between the
series `2 + 6 + 10 + · · ·` and `1 + 6 + 11 + · · ·`?
NDA Paper 1 2011

(A)

`180`

(B)

`186`

(C)

`196`

(D)

`206`

Solution:

Let the series,

`S_1 = 2 + (6) + 10 + 14 + 18 + 22 + (26) + 30 + 34 + 38 + 42`

` + (46) + ... `

and `S_2 =1 + (6) + 11+ 16+21+ (26) + 31+ 36+ 41+ (46) + ......`

The number sequence of common terms in `S_1`,

`S_1' = 2 + 7 + 12 + ...`

and number sequence of common terms in `S_2`,

`S_2' = 2 + 6 + 10 + ...`

Now, we find the `10`th term in both `S_1`' and `S_2`' ,

for `S_1' T_(10) = 2 + (10 - 1) · 5 = 2 + 45 = 47`

and for `S_2' T_(10) = 2 + (10 - 1) · 4 = 2 + 36 = 38`

So, the `47`th term in `S_1` and `38`.th term in `S_2` are the `10` th common

term in both series.

For `S_1, T_(47) = 2 + (47- 1) xx 4 = 2 + 46 xx 4 = 186`

and for `S_2, T_(38) = 1 + (38- 1) xx 5 = 1 + 37 xx 5 = 186`

Correct Answer is `=>` (B) `186`

Problem On HP

`T_n=1/(a+(n-1)d)`

Q 2763080845

The sum of the roots of the equation `x^2 + bx + c = 0` (where `b` and `c` are non-zero) is equal to the sum of the reciprocls of their squares. Then `1/c , b , c/b` are in
NDA Paper 1 2017

(A)

(B)

(C)

(D)

None of the above

Solution:

Let roots are `alpha, beta`

`alpha+ beta = 1/(alpha^2) + 1/(beta^2)`

`alpha+ beta =(alpha ^2 + beta^2)/((alpha beta)^2)`

`=((alpha+ beta)^2 - 2 alpha beta)/((alpha beta)^2)`

`= (b^2- 2c)/(c^2)`

`b^2 +b c^2= 2c`

`b= (2(1/c) (c/b))/((1/c )+(c/b))`

means `1/c , b , c/b` are in H.P

Correct Answer is `=>` (C) HP

Q 2317878780

If for positive real numbers `x,y` and `z` the numbers
`x + y, 2y` and `y + z` are in harmonic progression,
then which one of the following is correct?
NDA Paper 1 2007

(A)

x, y and z are in geometric progression

(B)

x, y and z are in arithmetic progression

(C)

x, y and z are in harmonic progression

(D)

None of the above

Solution:

Since, `x + y, 2 y` and `y + z` are in harmonic

progression.

`2 y = (2(x - y)(y + z))/(x+y+y+z)`

`=> 2 y(x + 2 y + z) = 2 (xy + xz + y^2 + yz)`

`=> 2xy + 4y^2 + 2yz = 2xy + 2xz + 2y^2 + 2yz`

` => 2y^2 =2xz`

` :. y^2 = xz`

So, `x, y` and `z` are in geometric progression.

Correct Answer is `=>` (A) x, y and z are in geometric progression

Q 2337478382

If `1/(b - c) + 1/(b - c) = 1/a + 1/c`, then `a, b` and `c` are in
NDA Paper 1 2008

(A)

(B)

(C)

(D)

None of these

Solution:

Let `a, b` and `c` be in HP.

`:. b = (2ac)/(a+c)`

`LHS = 1/(b - c) + 1/( b - a) = 1/((2ac)/(a + c) - c) + 1/((2ac)/(a + c) - a)`

` = (a+c )/( 2ac - ac - c^2) + ( a + c)/(2ac - a^2 - ac)`

` = (a+c) /(ac - c^2) + (a + c)/(ac-a^2)`

`= (a+ c) ( (ac - a^2 + ac - c^2)/(ac ( a - c)(c - a)))`

` = ((a+ c)(2ac - a^2 - c^2))/(-ac (c - a )^2)`

` = ( (a+c)(c - a)^2)/(-ac (c - a )^2) = (a+c)/(ac)`

`= a/(ac) + c/(ac) = 1/c + 1/a = RHS`

Hence, `a, b` and care in `HP`.

Correct Answer is `=>` (B) HP

Q 2367445385

If `1//4, 1//x` and `1//10` are in HP, then what is the
value of `x`?
NDA Paper 1 2012

(A)

`5`

(B)

`6`

(C)

`7`

(D)

`8`

Solution:

Given that, `1//4, 1//x` and `1//10` are in HP.

`=> 4, x` and `10` are in `AP`.

`:.` Arithmetic mean `x = (4 + 10)/2 = (14)/2 = 7 `

Correct Answer is `=>` (C) `7`

Q 2387534487

If the positive integers `a, b, c` and `d` are in `AP`, then
the numbers `abc, abd, acd ` and `bcd` are in
NDA Paper 1 2013

(A)

`HP`

(B)

`AP`

(C)

`GP`

(D)

None of these

Solution:

Given that `a, b, c` and dare in `AP`.

`=> 1/a, 1/b , 1/c` and `1/d` are in `HP`.

`=> bcd, acd, abd` and `abc` are in `HP`.

Hence, `abc, abd, acd` and `bcd` are in `HP`.

Correct Answer is `=>` (A) `HP`

Q 2855567464

If first three terms of sequence `1/16 , a , b , 1/6` are in geometric series and the last three terms are in harmonic series. then the values of a and b will he

(A)

` a = -1/4 , b = 1`

(B)

`a = 1/12 , b = 1/9`

(C)

Both (a) and (b)

(D)

None of these

Solution:

`because 1/16 , a , b ` are in GP

`=> a^2 = 1/16 xx b => b = 16 a^2` ............(i)

Also `a , b , 1/6` are in HP

`1/a , 1/b , 6 ` are in AP `=> 2/b = 6 +1/a`

`=> 2/(16 a^2) = ( 6a+1)/a => 8a ( 6a+1) = 1`

`=> 48a^2+8a-1 = 0`

`=> a = (-1)/4 , 1/12`

`therefore a = -1/4` and `a = 1/12` and `b = 1/9`

Correct Answer is `=>` (C) Both (a) and (b)

Q 2406745678

Let `a_1, a_2, .... , a_10` be in AP and `h_1, h_2, ....., h_10` be
in HP. If `a_1 = h_1 = 2` and `a_ 10 = h_10 = 3`. Then, `a_4h_7` is
UPSEE 2010

(A)

`2`

(B)

`3`

(C)

`5`

(D)

`6`

Solution:

Let `d` be the common difference of AP.

Then `a_10 =3`

`=> a_1+9 d=3`

`=> 2+9d=3`

`=> d=1//9`

`:. a_4 =a_1+3d =2+1/3=7/3`

Let `D` be the common difference of

`1/h_1 , 1/h_2 ,.......1/h_10 ` then

`h_10 =3=> 1/h_10 =1/3`

`=> 1/2+9 D=1/3`

`9 D=-1/6`

`=> D=-1/54`

`:. 1/h_7 =1/h_1+6D=1/2-1/9`

`=> h_7 =18/7`

`:. a_4 h_7 =7/3 xx 18/7 =6`

Correct Answer is `=>` (D) `6`

Q 1685623567

If `a_1,a_2,a_3, ... ,a_n` are in HP, then the expression
`a_1 a_2 + a_2a_3 + ... a_(n-1) a_n` is equal to
BITSAT 2016

(A)

`n(a_1 - a_n)`

(B)

`(n-1) (a_1 -a_n)`

(C)

`na_1a_n`

(D)

`(n-1)a_1a_n`

Solution:

Since, `a_1, a_2, a_3 ..... ,a_n` are in `HP.`

` :. 1/a_1 , 1/a_2 , 1/a_3 ,................,1/a_n` are in `AP.`

Let d be the common difference of the `AP` then

` 1/a_2 - 1/a_1 = d, 1/a_3 - 1/a_2 = d,.......1/a_n - 1/a_(n-1) = d`

` => a_1 - a_2 = d (a_1 a_2), a_2 - a_3 = d (a_2 a_3),`

`a_3 - a_4 = d (a_3 a_4), .... ' a_(n-1) -a_n = d (a_(n-1) a_n)`

` => (a_1 - a_2) + (a_2 - a_3) + .... + (a_(n-1) -a_n)`

`= d (a_1 a_2 + a_2 a_3 + .... + a_(n_1) a_n)`

` => a_1 - a_n = d (a_1 a_2 + a_2 a_3 + ... + a_(n-1) a_n)` ... (i)

Now,

`1/a_1 , 1/a_2 , 1/a_3 ,................1/a_n` are in `AP` with common ditference `d.`

` => 1/a_n = 1/a_1 + (n-1)d`

` => 1/a_n - 1/a_1 = (n-1)d`

` => (a_1 - a_n)/(a_1a_n) = (n-1) d (a_1a_n)` ..........(ii)

From Eqs. (i) and (ii), we get

`(n -1)d(a_1 a_n) = d(a_1 a_2 + a_2 a_3 + ... + a_(n-1) a_n)`

` => (n -1) a_1 a_n= a_1 a_2 + a_2 a_3 + ... + a_(n-1) a_n`

Correct Answer is `=>` (D) `(n-1)a_1a_n`

Q 2337156982

If `x^2 , y^2` and `z^2` are in `AP`, then `y + z, z + x, x + y`
are in
NDA Paper 1 2009

(A)

(B)

(C)

(D)

None of these

Solution:

Since, `2y^2 = x^2 + z^2 (∵ x^2, y^2` and `z^2` are in AP)

Let `y + z, z + x` and `x + y` are in HP.

`:. z + x = (2 (y + z)(x + y))/( y + z + x + y) => z + x = (2 (y + z)(x + y))/(2y + z + x)`

` = 2 y z + z^2 + z x + 2 x y + x z + x^2`

` =2yx + 2y^2 + 2zx + 2yz`

`=> z^2 + x^2 =2 y^2`

Hence, `x^2 , y^2` and `z^2` are in `AP`.

Correct Answer is `=>` (A) AP

Q 2327145981

If ` 1/( b - a) + 1/( b - c) = 1/a + 1/c` , then `a, b` and `c` is equal to
NDA Paper 1 2011

(A)

(B)

(C)

(D)

None of these

Solution:

Let `a, b` and `c` be in HP.

`:. b = (2ac)/(a+c)`

Now, `1/(b - a) + 1/(b - c) = 1/((2ac)/(a + c) - a) + 1/((2ac)/(a + c) - c)`

` = 1/(a ((2 c - a - c)/(a + c))) + 1/(c((2 a - a - c)/(a + c)))`

`= (a+c)/(a( c - a)) + (a+ c)/(c ( a - c))`

` = ((a+c)/( c-a)) ( 1/a - 1/c)`

` = ( a+ c)/( c - a) xx ( c -a)/(ca) = (a + c)/(ca) = 1/a + 1/c`

Hence, `a, b` and `c` are in `HP`.

Correct Answer is `=>` (C) HP

Q 2387345287

What does the series `1 + 1/sqrt(3) + 3 + 1/(3sqrt3) + ...`
represent ?
NDA Paper 1 2012

(A)

(B)

(C)

(D)

None of these

Solution:

Given series is

`1 + 1/sqrt(3) + 3 + 1/(3sqrt3) + ...`

Here, between each two consecutive terms, no common

difference and common ratio are form.

Hence, the given series does not form any series.

Correct Answer is `=>` (D) None of these

Problems on insertion of AM, GM, HM and their relation

`AM ge GM ge HM`

`(GM)^2 = AM . HM`

Q 2377678586

If the `n`th term of an arithmetic progression is
`3n + 7`, then what is the sum of its first `50` terms?
NDA Paper 1 2007

(A)

`3925`

(B)

`4100`

(C)

`4175`

(D)

`8200`

Solution:

Given that, `T_n = 3n + 7`

`S_n = sum T_n = sum (3n + 7) = 3 sum n + 7 sum 1`

`= (3n(n + 1))/2 + 7n = n [ (3n + 3 + 14)/2]`

`= n [ (3n - 17)/2 ]`

Now, sum of `50` terms `= S_(50) = 50 [ (3 xx 50 + 17)/2 ]`

`= 50 [ (167)/2] = 25 xx 167 = 4175`

Correct Answer is `=>` (C) `4175`

Q 2367867785

If the AM and GM of two numbers are `5` and `4`
respectively, then what is the HM of those
numbers?
NDA Paper 1 2008

(A)

`5/4`

(B)

`(16)/5`

(C)

`9/2`

(D)

`9`

Solution:

We know that, `HM = (GM)^2/(AM)`

`= (16)/5`

Correct Answer is `=>` (B) `(16)/5`

Q 2347756683

If the `AM` and `HM` of two numbers be `27` and `12`
respectively, then what is their `GM`?
NDA Paper 1 2010

(A)

`12`

(B)

`18`

(C)

`24`

(D)

`27`

Solution:

` ∵ AM= 27, HM = 12`

and we know that,

`(GM)^2 = (AM) (HM) = 27 xx 12`

`=> GM = 3 xx 3 xx 2 = 18`

Correct Answer is `=>` (B) `18`

Q 2387245187

The geometric mean and harmonic mean of two
non-negative observations are `10` and `8`,
respectively. Then, what is the arithmetic mean of
the observations?
NDA Paper 1 2012

(A)

`4`

(B)

`9`

(C)

`12.5`

(D)

`25`

Solution:

Given that, Geometric mean `(G)= 10`

and Harmonic mean `(H) = 8`

Let A be the arithmetic mean.

Then, `G^2 = AH = A = G^2/H`.

`=> A = (10)^2/8 = (100)/8 = 12.5`

Correct Answer is `=>` (C) `12.5`

Q 2845167963

Two AMs `A_1` and `A_2` two GM's `G_1` and `G_2` and two HM's `H_1` and `H_2` are inserted between any two numbers, then `H_1^(-1)+H_2^(-1)` is equal to

(A)

`A_1^(-1)+A_2^(-1)`

(B)

`G_1^(-1)+G_2^(-1)`

(C)

`(G_1 G_2)/(A_1+A_2)`

(D)

`(A_1+A_2)/(G_1 G_2)`

Solution:

Sum of nM's = n `xx` single AM

`therefore A_1+A_2 = 2 ((a+b)/2) = a+b ` ............(i)

Product of n GM's = (single GM)`text()^n`

`G_1 G_2 = ( sqrt(ab))^2 = ab` ..............(ii)

`1/a , 1/H_1 , 1/H_2 , 1/b` are in AP

`therefore 1/H_1+1/H_2 = 1/a+1/b = ( a+b)/(ab)` [from Eq. (i)]

`=> H_1^(-1) +H_2^(-1) = ( A_1+A_2)/(G_1 G_2)` [from Eqs. (i) and (ii)]

Correct Answer is `=>` (D) `(A_1+A_2)/(G_1 G_2)`

Q 2367845785

If the arithmetic and geometric means of two
numbers are `10, 8` respectively, then one number
exceeds the other number by
NDA Paper 1 2011

(A)

`8`

(B)

`10`

(C)

`12`

(D)

`16`

Correct Answer is `=>` (C) `12`

Q 1669445315

Which one of the following measures of central
tendency is used in construction of index
numbers?
NDA Paper 1 2015

(A)

Harmonic mean

(B)

Geometric mean

(C)

Median

(D)

Mode

Solution:

Harmonic mean is used in construction of index

numbers.

Correct Answer is `=>` (A) Harmonic mean

Q 2367034885

The harmonic mean `(H)` of two numbers is `4` and
the arithmetic mean `(A)` and geometric mean `(G)`
satisfy the equation `2A + G^2 = 27`. The two
numbers are
NDA Paper 1 2013

(A)

`6` and `3`

(B)

`9` and `5`

(C)

`12 ` and `7`

(D)

`3` and `1`

Solution:

Given that.

Harmonic mean (H) of two numbers `= 4`

Let the two numbers be `a` and `b`.

Also, given that

` 2 A + G^2 = 27`... (i)

We know that, relation between arithmetic mean `(A)`, geometric

mean (G) and harmonic mean (H) is

` G^2 = AH`

which satisfy the Eq. (i).

`:. 2 A+ AH = 27 => 2 A + A · 4 = 27`

`=> 6A = 27`

` :. A = 9/2` ... (ii)

Arithmetic mean 'A' of two numbers a and b is

` (a + b )/2 = A = 9/2`, [from Eq. (ii)]

` => a + b = 9` ..(iii)

` ∵ H = (2ab)/(a + b) = (2ab)/9 = 4 => ab = 18` .....(iv)

`:.` We have, `(a - b)^2 = (a + b)^2 - 4ab = (9)^2 - 4·18 = 81 - 72 = 9`

`=> a - b = ± 3` ... (v)

From Eqs. (iii) and (v),

Case I `a + b = 9` and `a - b = 3`

`=> 2a = 12 => a = 6`

and `b = 3`

Case II `a + b = 9` and `a - b = - 3`

`=> 2a = 6 => a = 3`

and `b = 6`

Hence, required numbers are `6` and `3` or `3` and `6`.

Correct Answer is `=>` (A) `6` and `3`

Sum upto `n` Terms of Special Series `sum n, sum n^2 , sum n^3`, etc

`sum n= (n(n+1))/2`

`sum n^2=(n(n+1)(2n+1))/6`

`sum n^3=[ (n(n+1))/2]^2=(sumn)^2`

`sum n^4 = (n(n+1)(2n+1)(3n^2+3n-1))/30`

Q 2347867783

If the `n`th term of an arithmetic progression is
`2n -1`, then what is the sum upto `n` terms?
NDA Paper 1 2008

(A)

`n^2`

(B)

`n^2 - 1`

(C)

`n^2 + 1`

(D)

` 1/2 n( n + 1)`

Solution:

Given, `a_n = 2n - 1`

`:. S_n = sum_( k = 1)^n a_k = sum_( k = 1)^n (2n -1)`

`= 2 sum_( k = 1)^n n -n`

`= 2. (n(n + 1))/2 - n = n^2 + n - n = n^2`

Correct Answer is `=>` (A) `n^2`

Q 2327134981

Consider the following statements
I. The sum of cubes of first `20` natural numbers
is `44400`.
II. The sum of squares of first `20` natural numbers
is `2870`.
Which of the above statements is/are correct?
NDA Paper 1 2012

(A)

Only I

(B)

Only I

(C)

Both I and II

(D)

Neither I nor II

Solution:

Since, the sum of cubes of first `n` natural numbers

`= [ (n(n + 1))/2]^2`

and the sum of squares of first `n` natural numbers

`= (n(n + 1)(2n + 1))/6`

`:.` The sum of cubes of first `20` natural numbers

` = [ (20(20 + 1))/2]^2 = ( (20 xx 21)/2)^2`

`= (10 xx 21)^2 = 44100`

and the sum of squares of first `20` natural numbers

` ( 20(20 + 1)(2 xx 20 + 1))/6`

` = (20 xx 21 xx 41)/6 = 2870`

Correct Answer is `=>` (B) Only I

Q 2222001831

The arithmetic mean of `1, 8, 27, 64` , ... upto n terms is
given by
NDA Paper 1 2015

(A)

` (n(n+1))/2`

(B)

` (n(n+1)^2)/2`

(C)

` (n(n+1)^2)/4`

(D)

` (n^2(n+1)^2)/4`

Solution:

Given, `1, 8, 27, 64`, ..... upto `n` terms

`= 1^3, 2^3, 3^ 3, 4^3 ,..........` upto `n` terms

`:. AM =(1^3+ 2^3+ 3^ 3+ 4^3 ..........+n^3)/n = [(n(n+1))/2]^2/n`

` = (n^2(n + 1)^2)/(4n) = ( n(n + 1)^2)/4`

`[ ∵ sum n^3 = 1^2 + 2^2 + ... + n^2 = {n/2 (n+ 1) }^2]`

Correct Answer is `=>` (C) ` (n(n+1)^2)/4`