Mathematics Previous Year Sequence And Series Questions For NDA

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Previous Year Sequence And Series Questions For NDA

Q 2753780644

What is the sum of the series `0· 3 + 0· 33 + 0· 333 + ... n` terms ?
NDA Paper 1 2017

(A)

`1/3 [ n - 1/9 (1 - 1/(10)^n ) ]`

(B)

`1/3 [ n - 2/9 (1 - 1/(10)^n ) ]`

(C)

`1/3 [ n - 1/3 (1 - 1/(10)^n ) ]`

(D)

`1/3 [ n - 1/9 (1 + 1/(10)^n ) ]`

Solution:

`.3 +.33 +.333+ ...............n ` terms

`=1/3 [.9 +.99 +.999+ .....................n]`

`= 1/3 [(1-.1 +(1-.01)+(1- .001)+.............]`

`=1/3 [(1+1+.......n " terms ") + (1/10 + 1/100 +1/1000+............ n " terms ")]`

`=1/3 [n-(1/10 (1- (1/10)^n))/((1-1/10))]`

`=1/3 [n-1/9 (1- 1/(10^n))]`

Correct Answer is `=>` (A) `1/3 [ n - 1/9 (1 - 1/(10)^n ) ]`

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 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 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 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 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 2711556420

The sixth term of an AP is 2 and its common difference is greater than 1.
What is the common difference of the AP so that the product of the first, fourth and fifth terms is greatest ?
NDA Paper 1 2016

(A)

8/5

(B)

9/5

(C)

(D)

11/5

Solution:

`a+5d'=2,d'> 1`

`p= a (a+3d') (d+4d')`

Put the value of a from given relation

`p= (2-5d') (2-2d) (2-d')`

`=+ 2 (1-d) (2-d) (2-5d')`

to be greatest of value `p`

`(dp)/(dd') =0`

`d/(d') [+2 (1-d') (2-d') (2-5d') ] = 0`

`(1-d') (2-d') 5 + (5 d' -2) (3-2d') =0`

`5d^2 -15 d+10 +10 d^2 - 19d + 6 = 0`

`15 d^2 -15 d +10 +10d^2 -19 d+6 =0`
`d= (+34 pm sqrt((34)^2-4 xx 15 xx 16))/(2 xx 15)`

`= (34 pm 14)/(30) = 48/30 , (20/30 (d < 1) )` rejected

`=8/5`

Correct Answer is `=>` (A) 8/5

Q 2711556420

The sixth term of an AP is 2 and its common difference is greater than 1.
What is the first term of the AP so that the product of the first , fourth and fifth terms is greatest ?
NDA Paper 1 2016

(A)

-4

(B)

-6

(C)

-8

(D)

-10

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

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 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 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 2741756623

The interior angles of a polygon of n sides are in AP. The smallest angle is 120° and the common difference is 5°.
What is the largest interior angle of the polygon ?
NDA Paper 1 2016

(A)

160° only

(B)

195° only

(C)

Either 160° or 195°

(D)

Neither 160° nor 195°

Correct Answer is `=>` (D) Neither 160° nor 195°

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 2331080822

If the value of the detenninant `[(a,1,1),(1,b,1),(1,1,c)] ` is positive where `a != b != c` then the value of ` abc`
NDA Paper 1 2015

(A)

cannot be less than 1

(B)

is greater than- 8

(C)

is less than- 8

(D)

must be greater than 8

Solution:

Let `Delta = | (a,1,1),(1,b,1),(1,1,c)| = a(bc -1) - 1 (c-1) + 1(1 -b)`

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

`because Delta > 0`

`:. abc - a - b - c + 2 > 0`

`=> abc+ 2 > a+ b + c => abc+ 2 > 3(abc) ^1/3`

`:. a != b != c`

`:.` `AM` of `a, b, c > GM` of `a, b, c`

`=> (a +b + c)/3 > (abc)^(1//3)`

Now , let ` x = (abc)^(1//3) ` , then we have

`x^3 + 2 > 3 x => x^3 - 3 x + 2 > 0`

`=> (x - 1 )^2 (x + 2) > 0`

`x+2 > 0` `[ because (x -1 )^2 > 0]`

`=> x > -2 => (abc)^(1//3) > -2`

`=> abc > -8`

Correct Answer is `=>` (B) is greater than- 8

Q 1608723608

What is the sum of the series

`0.5 + 0.55 + 0.555 + ....` to `n` terms?
NDA Paper 1 2015

(A)

`5/9 [ n - 2/9 (1- 1/(10)^n)]`

(B)

`1/9 [ 5 - 2/9 (1- 1/(10)^n)]`

(C)

`1/9 [ n - 5/9 (1- 1/(10)^n)]`

(D)

`5/9 [ n - 1/9 (1- 1/(10)^n)]`

Solution:

`0.5 + 0.55 + 0.555 + ...` to `n` terms

`5/(10) + (55)/(100) +( 555)/(1000) + .....` upto `n` terms

` = 5/(10) [1 + (11)/(10) + (111)/ (100) + ..n]`terms

` = 5/(10) xx 1/9 [ 9 + (99)/(10) + (999)/ (100) + ... ` upto `n` terms]

` = 5/(90) [ (10 -1 ) + ( (10^2 -1))/(10) + ( (10^3 -1))/(10^2) ..... n` terms

` = 5/(10) [ (10^2)/(10) + (10^3)/(10^2) + ....` upto `n` terms]

` + 5/(90) [ -1 - 1/(10) - 1/(10)^(-2) - 1/(10)^(-3) - ....` upto `n` terms]

` = 5/(90) [ 10 + 10 + 10 + ....` upto `n` terms]

`+ 5/(90) (-1) [-1 1/(10) - 1/(10^2) _-1/(10^3) - ....` upto `n` terms]

`= 5/(90) xx 10n - 5/(90) [ ( 1-(-10^(-1))n)/(1- 1/(10))]`

` = (5n)/9 - 5/(90) xx(10)/9 [ (10^n -1)/(10^n)] = 5/9 [ n - 1/9 ( 1 - 1/(10^n) )]`

Correct Answer is `=>` (D) `5/9 [ n - 1/9 (1- 1/(10)^n)]`

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 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 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 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 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 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`

Q 1629067811

What is `lim_(n -> oo) ( 1+2+3+ .... + n)/(1^2 + 2^2 + 3^2 + ... + n^2)` equal to?

NDA Paper 1 2014

(A)

`5`

(B)

`2`

(C)

`1`

(D)

`0`

Solution:

`lim_(n -> oo) ( 1+2+3+ .... + n)/(1^2 + 2^2 + 3^2 + ... + n^2)`

`= lim_(n -> oo) ( (n(n+1))/2)/( (n(n+1)(2n+1))/6`

` [ ∵1 + 2 + 3 + .... + n = (n(n+1))/2`

and `1^2 + 2^2 + 3^2 + ... + n^2 = (n(n+1)(2n+1))/6]`

` = lim_(n -> oo) 3/(2n +1) = 0`

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

Q 1609067818

Consider the given information
Let `S_n` denotes the sum of first `n` terms of an `AP` and
`3S_n = 2.S_(2n)`

What is `S_(3n) : S_(2n)` equal to?
NDA Paper 1 2014

(A)

`2 : 1`

(B)

`3 : 1`

(C)

`4 : 1`

(D)

`5: 1`

Solution:

We have, `S_n` = Sum of first `n` terms of an `AP`

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

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

and `S_(3n) = (3n)/2 [2a + (3n -1)d]`

Now, `3S_n = 2S_(2n)`

`=> 3 (n/2) [2a + (n -1)d] = 2 (2n/2) [2a + (2n -1)d]`

` => 2a = d (n + 1)`

`:. S_ n = n/2 [d (n + 1) + d (n - 1)]`

` = n/2 (2nd) = n^2 d`

Now, `S_(2n) = n [d(n + 1 + 2n -1)] = 3n^2 d`

and `S_(3n) = (3n)/2 [d(n + 1 + 3n -1)]= 6n^2 d`

`:. (S_(3n))/(S_n) = (6n^2 d)/(n^2d) = 6 : 1`

` (S_(3n))/(S_(2n)) = (6n^2 d)/(3n^2d) = 2 : 1`

Correct Answer is `=>` (A) `2 : 1`

Q 1702380238

The sum of an infinite `GP` is `x` and the common ratio `r` is
such that ` | r | < 1`. If the first term of the GP is `2`, then
which one of the following is correct?
NDA Paper 1 2014

(A)

`-1 < x < 1`

(B)

`- oo < x < 1`

(C)

`1 < x < oo`

(D)

None of these

Solution:

Given that,

The sum of an infintie `GP = x`

`=> a/(1-r) = x`,

where, `a =` first term and `r` = common ratio

` => 2/(1-r) = x` ...........(1)

`(∵` given that, `a = 2` and `| r | < 1)`

`∵ | r | < 1`

`=> - 1 < r < 1 `

` => 1 > - f > - 1 `

`=> 1 + 1 > 1 - r > 1 - 1 `

` => 0 < 1 - r < 2 `

` => (1- r) < 2`

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

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

` => x > 1` [from Eq. (1)]

Hence, `x in (1, oo)` i.e., `1 < x < oo`.

Correct Answer is `=>` (C) `1 < x < oo`

Q 1722334231

Let `f(x) = ax^2 + bx + c` such that `f(1) = f( -1)` and `a, b, c`
are in arithmetic progression.

What is the value of `b?`
NDA Paper 1 2014

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

Cannot be determined due to insufficient data

Solution:

We have, `f(x) = ax^2 + bx + c`

`:. f(1) = a + b + c`

and `f(-1) = a - b + c`

`:. f(1) = f(-1)`

`=> a + b + c = a - b + c => b = 0`

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

Q 1732334232

Let `f(x) = ax^2 + bx + c` such that `f(1) = f( -1)` and `a, b, c`
are in arithmetic progression.

`f ' (a) , f ' (b)` and `f ' (c)` are in
NDA Paper 1 2014

(A)

`AP`

(B)

`GP`

(C)

`HP`

(D)

Arithmetico-geometric progression

Solution:

We have, `f' (x) = 2ax + b'`

`:. f'(a) = 2a^2, f' (b)= 2ab = 0`

and `f' (c) = 2ac quad ( ∵ b = 0 )`

`:. f ' (a) = 2 a^2`

` f ' (b) = 0`

and `f'(c) = -2a^2 quad ( ∵ 2b = a + c => c = -a )`

Hence, `f'(a), f'(b)` and `f'(c)` are in `AP`.

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

Q 1742334233

Let `f(x) = ax^2 + bx + c` such that `f(1) = f( -1)` and `a, b, c`
are in arithmetic progression.

`f '' (a), f '' (b)` and `f'' (c)` are
NDA Paper 1 2014

(A)

in `AP` only

(B)

in `GP` only

(C)

in both `AP` and `GP`

(D)

neither in `AP` nor in `GP`

Solution:

` f '' (x) = 2a`

`:. f '' (a) = f '' (b) = f '' (c)`

Hence, `f '' (a), f '' (b)` and `f '' (c)` are in both `AP` and `GP`.

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

Q 1783178047

The sum of the series formed by the sequence `3, sqrt(3) , 1, ...`
upto infinity is
NDA Paper 1 2014

(A)

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

(B)

` (3sqrt(3) (sqrt(3) - 1))/2`

(C)

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

(D)

`(3(sqrt(3) - 1))/2`

Solution:

Given series, `3, sqrt(3) , 1, .... , oo`

and the series form an infinite `GP`.

whose first term `(a) = 3`

and common ratio `(r) = sqrt(3)/3 = 1/sqrt(3)`

`:.` Sum of infinite `GP , S_(oo) = a /(1-r) = 3/(1 - 1/sqrt(3))`

` = (3sqrt(3))/sqrt((3) - 1) . (sqrt(3) + 1)/(sqrt(3) + 1)`

` = (3sqrt(3) ( sqrt(3) + 1) )/ (3-1) `

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

Correct Answer is `=>` (A) ` (3sqrt(3) (sqrt(3) + 1))/2`

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 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 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 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 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`

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 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 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 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 the common difference'?
NDA Paper 1 2012

(A)

`1`

(B)

`1`

(C)

`3`

(D)

`4`

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

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 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`

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 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 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

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 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 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 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 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 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`