Create New Account
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

Arithmetic Progressions

    Sum of First n Terms of an AP

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

10th class > Arithmetic Progressions > Sum of First n Terms of an AP

Sum of First n Terms of an AP

Let us consider the situation again given in the earlier section, in which Shakila put ₹ 100 into her daughter’s money box when she was one year old, ₹ 150 on her second birthday, ₹ 200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old?

Here, the amount of money (in ₹) 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 .

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 = 100x1012 =

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 + ) + (a + ) + . . . + [a + (n – 1) ] (1)

Rewriting the terms in reverse order, we have

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

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

So, S = n2 [2a + (n – 1) d]

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

S = n2 [2a + (n – 1)d]

We can also write this as S = n2 [a + a + (n – 1) d ] i.e., S = n2(a + an) (3)

Now, if there are only n terms in an AP, then an = l, the last term.

From (3), we see that S = n2(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 = , d = and n = .

Using the formula : S = n2 [2a + (n – 1) d ]

We have S = 212 [2 x 100 + (21-1)50] = 212 [200 + 1000] = 212 x 1200 =

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 Sn in place of S to denote the sum of first n terms of the AP. We write S20 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., an = Sn – Sn−1.

Let us consider some examples.

Example 11

Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .

Solution :

Here, a = , d = 3 – 8 = , n = .

We know that: S = n2 [2a + (n – 1) d]

Putting values, we get:

S = 222 [16 + 21(-5)] = 11(16 – 105) = 11 × =

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

Example 12

Example 12 :If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Solution :

Here, S14 = , n = , a = .

As Sn= n2 [2a + (n – 1) d ]

1050 = 142 [20 + 13d] = + 91d

910 = d

d =

Therefore, a20 = + ( – ) × =

i.e. 20th term is 200.

Example 13 : How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?

Solution : Here, a = , d = 21 – 24 = , Sn = .

We need to find n.

We know that: Sn = n2 [2a + (n – 1) d ]

78 = n2 ( + (n-1)(-3)) = n2( - 3n)

3n2– n + = 0

n2 – 17n + = 0

(n – )(n – ) = 0

n = or

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.

Example 14

Find the sum of :

(i) the first 1000 positive integers

(ii) the first n positive integers

Solution :

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

Using the formula Sn = n2(a+l) for the sum of the first n terms of an AP, we have:

S1000 = 10002 (1 + 1000) = × =

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

(ii) Let Sn = 1 + 2 + 3 + . . . + n

Here a = and the last term l is .

Therefore, Sn =

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

Sn = nn+12

Example 15

Example 15: Find the sum of first 24 terms of the list of numbers whose nth term is given by an = 3 + 2n

Solution :

As an = 3 + 2n,

so, a1 = 3 + 2 =

a2 = 3 + 2 × 2 =

a3 = 3 + 2 × 3 = and so on .........

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

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

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

To find S24, we have n = , a = , d = .

Therefore, S24 = 242 [2 x 5 + (24-1) x 2] = 12 [10 + 46] =

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

Example 16 : 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

Find the answer

  • 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 an.
  • Here, we have: a3 = while a7 =
  • Using the nth term formula: an=a+n−1d, we substitute the values to get the required equations.
  • So, a3 we get: a + = 600 while for a7 we have: a + =
  • Solving both the equations, We get : a = and d =
  • Thus, production of TV sets in the first year is .
  • Production of TV sets in the 10th year is .
  • Total production of TV sets in first 7 years is .
  • We have found the answer.