Question

If 7 th term of an AP is 40 then find the sum of its first 13 term.

Answer 1

Answer:

13th term = 520

Step-by-step explanation:

7th term = 40

AP⇒ nth term=$a_{n}=a+(n-1)d$

$a_{n}=40$

$n=7$

$a+(7-1)d=40$

$40=a+6d$------(i)

$s_{n}=\frac{n}{2}[2a+(n-1)d]$

$s_{13}=\frac{13}{2}[2a+(13-1)d]$

$s_{13}=\frac{13}{2}[2a+12d]$

$s_{13}=\frac{13}{2}×2[a+6d]$  [here,2 and 2 gets cancel]  

$s_{13}=13×40$    [∴a+6d=40]

$→s_{13}=520$

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