Question

the 7th term of an ap is 40 then the sum of first 13 terms is

Answer 1

Sum of n terms =n/2(1st term+ nth term). Let the first term of an A.P. be a and common difference is d. So the 7th term of the A.P. is 40=> a+6d =40. Sum of first 13 terms = 13/2(a +(a+12d))= 13/2(2a+12d) =13/2(2)(a+6d) =13(a+6d) =13(40)=520 . Hope it helps you... Mark me as Brainliest.

Answer 2

Answer:

Sum of 13 terms is 520.

Step-by-step explanation:

Given: seventh term of an AP = 40

To find: Sum of first 13 terms.

Let, a be the first term and d be the common difference.

We know that,

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

$a_7=40$

a + ( 7 - 1 )d = 40

a + 6d = 40

Also,

$S_n=\frac{n}{2}(2a+(n-1)d)$

$S_{13}=\frac{13}{2}(2a+(13-1)d)=\frac{13}{2}(2a+12d)=\frac{13}{2}\times2(a+6d)=13\times40=520$

Therefore, Sum of 13 terms is 520.