Sum

The *n*th term of an AP is given by (−4*n* + 15). Find the sum of first 20 terms of this AP?

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#### Solution

Given, *a*_{n} = − 4*n* + 15

∴ *a*_{1} = − 4 × 1 + 15 = − 4 + 15 = 11

*a*_{2} = − 4 × 2 + 15 = − 8 + 15 = 7

*a*_{3} = − 4 × 3 + 15 = − 12 + 15 = 3

*a*_{4} = − 4 × 4 + 15 = − 16 + 15 = −1

It can be observed that

*a*_{2} − *a*_{1} = 7 − 11 = −4

*a*_{3} − *a*_{2} = 3 − 7 = −4

*a*_{4} − *a*_{3} = − 1 − 3 = −4

i.e., *a*_{k}_{ + 1} − *a*_{k} is same every time. Therefore, this is an A.P. with common difference as

−4 and first term as 11.

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

`S_20=20/2[2(11)+(20-1)(-4)]`

`=10[22+19(-4)]`

`=10(20-76)`

`=10(-54)`

`=-540`

Concept: Sum of First n Terms of an AP

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