Question

In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?

Answer 1

In the given problem, let us take the first term as a and the common difference d

Here, we are given that,

`a_5 = 30`    ....(1)

`a_12 = 65` .....(2)

Also, we know

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

For the 5th term (n = 5)

`a_5 = a + (5 - 1)d`

30 = a + 4d       (Using 1)

a = 30 - 4d ....(3)

Similarly for the 12 th term (n = 12)

`a_12 = a + (12 - 1)d`

65 = a + 11d     (Using 2)

a = 65 - 11d....(4)

Substracting (3) from (4) we get

a - a = (65 - 11d)-(30 - 4d)

0 = 65 - 11d - 30 + 4d

0 = 35 - 7d

7d = 35

d = 5

Now, to find a, we substitute the value of d in (4),

a = 30 - 4(5)

a = 30 - 20

a = 10

So for the given A.P d = 5 and a = 10

So to find the sum of first 20 terms of this A.P. we use the following formula for the sum of n terms of an A.P

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

Where a = first term for the given A.P

d= common difference of the given A.P

n = number of terms

So using the formula for n = 20 we get

`S_20 = 20/2 [2(10) + (20 - 1)(5)]`

= (10)[20 + (19)(5)]

= (10)[20 + 95]

= (10)[115]

= 1150

Therefore, the sum of first 20 terms for the given A.P. is `S_20 = 1150`