Please Wait... While Loading Full Video#### Class 10 Chapter 5 - ARITHMETIC PROGRESSIONS

♦ Sum of First n Terms of an AP

● Let us consider the situation again given in Section 5.1 in which Shakila put Rs. 100 into her daughter’s money box when she was one year old, Rs. 150 on her second birthday, Rs. 200 on her third birthday and will continue in the same way.

● Here, the amount of money (in Rs .) put in the money box on her first, second, third, fourth . . . birthday were respectively `100, 150, 200, 250, . . .` till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see.

● We consider the problem given to Gauss, to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote :

`S = 1 + 2 + 3 + . . . + 99 + 100`

And then, reversed the numbers to write

`S = 100 + 99 + . . . + 3 + 2 + 1`

Adding these two, he got

`2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100)`

`= 101 + 101 + . . . + 101 + 101` (100 times)

So, `S = (100 xx 101)/2 = 5050` , i.e., the sum ` = 5050` .

● We will now use the same technique to find the sum of the first n terms of an AP : `a, a + d, a + 2d, . . .`

● The nth term of this AP is `a + (n – 1) d`. Let S denote the sum of the first n terms of the AP. We have

`S = a + (a + d ) + (a + 2d) + . . . + [a + (n – 1) d ] `...............(1)

● Rewriting the terms in reverse order, we have

`S = [a + (n – 1) d] + [a + (n – 2) d ] + . . . + (a + d) + a` ...........(2)

On adding (1) and (2), term-wise. we get

`2S = ( [ 2a + ( n -1) d ] + [ 2a + ( n-1) d] + ..... + [2a + (n -1) d ] + [2a + ( n-1) d] )/( n text ( times ) )`

or, `2S = n [2a + (n – 1) d ] `(Since, there are n terms)

or, `S = n/2 [2a + (n – 1) d ]`

● So, the sum of the first n terms of an AP is given by

`color{orange}{S = n/2 [ 2a + ( n-1 ) d ]}`

We can also write this as `S = n/2 [ a+a + (n-1 ) d ]`

i.e., `S = n/2 (a + a_n )` ......... (3)

● Now, if there are only n terms in an AP, then `a_n = l`, the last term. From (3), we see that

`color{orange}{S = n/2 (a +l )}` ............(4)

● This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.

● Now we return to the question that was posed to us in the beginning. The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,

. . ., were `100, 150, 200, 250, . . .,` respectively.

● This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.

Here, a = 100, d = 50 and n = 21. Using the formula :

● `S= n/2 [2a + (n -1 ) d ]` ,

we have `S = (21)/2 [2 xx 100 + ( 21 -1 ) xx 50 ] = 21/2 [200 + 1000 ]`

` = 21/2 xx 1200 = 12600`

● So, the amount of money collected on her 21st birthday is Rs. 12600.

Hasn’t the use of the formula made it much easier to solve the problem?

● We also use `S_n` in place of S to denote the sum of first n terms of the AP. We write `S_(20)` to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth

`"Remark :"` The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., `color{orange}{a_n = S_n – S_(n – 1)}`.

● Here, the amount of money (in Rs .) put in the money box on her first, second, third, fourth . . . birthday were respectively `100, 150, 200, 250, . . .` till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see.

● We consider the problem given to Gauss, to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote :

`S = 1 + 2 + 3 + . . . + 99 + 100`

And then, reversed the numbers to write

`S = 100 + 99 + . . . + 3 + 2 + 1`

Adding these two, he got

`2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100)`

`= 101 + 101 + . . . + 101 + 101` (100 times)

So, `S = (100 xx 101)/2 = 5050` , i.e., the sum ` = 5050` .

● We will now use the same technique to find the sum of the first n terms of an AP : `a, a + d, a + 2d, . . .`

● The nth term of this AP is `a + (n – 1) d`. Let S denote the sum of the first n terms of the AP. We have

`S = a + (a + d ) + (a + 2d) + . . . + [a + (n – 1) d ] `...............(1)

● Rewriting the terms in reverse order, we have

`S = [a + (n – 1) d] + [a + (n – 2) d ] + . . . + (a + d) + a` ...........(2)

On adding (1) and (2), term-wise. we get

`2S = ( [ 2a + ( n -1) d ] + [ 2a + ( n-1) d] + ..... + [2a + (n -1) d ] + [2a + ( n-1) d] )/( n text ( times ) )`

or, `2S = n [2a + (n – 1) d ] `(Since, there are n terms)

or, `S = n/2 [2a + (n – 1) d ]`

● So, the sum of the first n terms of an AP is given by

`color{orange}{S = n/2 [ 2a + ( n-1 ) d ]}`

We can also write this as `S = n/2 [ a+a + (n-1 ) d ]`

i.e., `S = n/2 (a + a_n )` ......... (3)

● Now, if there are only n terms in an AP, then `a_n = l`, the last term. From (3), we see that

`color{orange}{S = n/2 (a +l )}` ............(4)

● This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.

● Now we return to the question that was posed to us in the beginning. The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,

. . ., were `100, 150, 200, 250, . . .,` respectively.

● This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.

Here, a = 100, d = 50 and n = 21. Using the formula :

● `S= n/2 [2a + (n -1 ) d ]` ,

we have `S = (21)/2 [2 xx 100 + ( 21 -1 ) xx 50 ] = 21/2 [200 + 1000 ]`

` = 21/2 xx 1200 = 12600`

● So, the amount of money collected on her 21st birthday is Rs. 12600.

Hasn’t the use of the formula made it much easier to solve the problem?

● We also use `S_n` in place of S to denote the sum of first n terms of the AP. We write `S_(20)` to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth

`"Remark :"` The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., `color{orange}{a_n = S_n – S_(n – 1)}`.

Q 3159280114

Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .

Class 10 Chapter 5 Example 11

Class 10 Chapter 5 Example 11

Here, `a = 8, d = 3 – 8 = –5, n = 22`.

We know that

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

Therefore , `S = 22/2 [16 +21 (-5) ] =11 (16 -105 ) =11 (-89) = -979`

So, the sum of the first 22 terms of the AP is – 979.

Q 3169280115

If the sum of the first 14 terms of an AP is 1050 and its first term is 10,

find the 20th term.

Class 10 Chapter 5 Example 12

find the 20th term.

Class 10 Chapter 5 Example 12

Here, `S_14 = 1050, n = 14, a = 10`.

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

so , `1050 = 14/2 [20 +13 d ] = 140 +91 d`

i.e., `910 = 91d`

or, `d = 10`

Therefore, `a_20 = 10 + (20 – 1) × 10 = 200`, i.e. 20th term is 200.

Q 3179280116

How many terms of the AP : 24, 21, 18, . . . must be taken so that their

sum is 78?

Class 10 Chapter 5 Example 13

sum is 78?

Class 10 Chapter 5 Example 13

Here, `a = 24, d = 21 – 24 = –3, S_n = 78`. We need to find n.

We know that `S_n = n/2 [2a + (n-1 ) d ]`

So, `78 = n/2 [48 + (n-1 ) (-3) ] = n/2 [51-3n ]`

or `3n^2 – 51n + 156 = 0`

or `n^2 – 17n + 52 = 0`

or `(n – 4)(n – 13) = 0`

or `n = 4` or `13`

Both values of n are admissible. So, the number of terms is either 4 or 13

Remarks :

1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.

2. Two answers are possible because the sum of the terms from 5th to 13th will be

zero. This is because a is positive and d is negative, so that some terms will be

positive and some others negative, and will cancel out each other.

Q 3189280117

Find the sum of :

(i) the first 1000 positive integers (ii) the first n positive integers

Class 10 Chapter 5 Example 14

(i) the first 1000 positive integers (ii) the first n positive integers

Class 10 Chapter 5 Example 14

(i) Let` S = 1 + 2 + 3 + . . . + 1000`

Using the formula `S_n=n/2 (a+l)` for the sum of the first n terms of an AP, we

have

`S_(1000)= 1000/2 (1+1000) =500 xx 1001 =500500`

So, the sum of the first 1000 positive integers is 500500.

(ii) Let `S_n = 1 + 2 + 3 + . . . + n`

Here a = 1 and the last term l is n.

Therefore, `S_n = (n (1+n) )/2 ` or `S_n= ( n (n+1 ) )/2`

So, the sum of first n positive integers is given by

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

Q 3109280118

Find the sum of first 24 terms of the list of numbers whose nth term is

given by

`a_n = 3 + 2n`

Class 10 Chapter 5 Example 15

given by

`a_n = 3 + 2n`

Class 10 Chapter 5 Example 15

As `a_n = 3 + 2n`,

so, `a_1 = 3 + 2 = 5`

`a_2 = 3 + 2 × 2 = 7`

`a_3 = 3 + 2 × 3 = 9`

.

.

.

.

List of numbers becomes `5, 7, 9, 11, . . .`

Here,` 7 – 5 = 9 – 7 = 11 – 9 = 2` and so on.

So, it forms an AP with common difference d = 2.

To find `S_24`, we have `n = 24, a = 5, d = 2`.

Therefore, `S_(24) = 24/2 [ 2 xx 5 + (24 -1 ) xx 2 ] =12 [10 +46 ] = 672`

So, sum of first 24 terms of the list of numbers is 672.

Q 3119280119

A manufacturer of TV sets produced 600 sets in the third year and 700

sets in the seventh year. Assuming that the production increases uniformly by a fixed

number every year, find :

(i) the production in the 1st year (ii) the production in the 10th year

(iii) the total production in first 7 years

Class 10 Chapter 5 Example 16

sets in the seventh year. Assuming that the production increases uniformly by a fixed

number every year, find :

(i) the production in the 1st year (ii) the production in the 10th year

(iii) the total production in first 7 years

Class 10 Chapter 5 Example 16

(i) Since the production increases uniformly by a fixed number every year,

the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.

Let us denote the number of TV sets manufactured in the nth year by `a_n`.

Then, `a_3 = 600` and `a_7 = 700`

or,` a + 2d = 600`

and `a + 6d = 700`

Solving these equations, we get `d = 25 `and `a = 550`.

Therefore, production of TV sets in the first year is `550`.

(ii) Now `a_10 = a + 9d = 550 + 9 × 25 = 775`

So, production of TV sets in the 10th year is `775`.

(iii) Also, `S_7 =7/2 [2 xx 550 + (7-1) xx 25 ]`

` = 7/2 [ 1100 +150 ] = 4375`

Thus, the total production of TV sets in first 7 years is` 4375`.