Question

d=5, s9=75, find n and a9

Answer 1

Given that common difference d=5

sum of 9 terms S9=75

S9 means n=9

So we will be going to use sum of n terms formula of arithmetic sequence.

Which is given by:

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

Plug the given values

$75=\frac{9}{2}(2a+(9-1)*5)$

$75=\frac{9}{2}(2a+(8)*5)$

$75=\frac{9}{2}(2a+40)$

$75=\frac{9}{2}2(a+20)$

$75=9(a+20)$

$\frac{75}{9}=a+20$

$\frac{25}{3}=a+20$

$\frac{25}{3}-20=a$

$\frac{25}{3}-20*\frac{3}{3}=a$

$\frac{25}{3}-\frac{60}{3}=a$

$\frac{25-60}{3}=a$

$\frac{-35}{3}=a$

Now we need to find a9 what is 9th term so we will use following formula

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

where $a=\frac{-35}{3}$, n=9, d=5

$a_9=\frac{-35}{3}+(9-1)*5$

$a_9=\frac{-35}{3}+8*5$

$a_9=\frac{-35}{3}+40$

$a_9=\frac{-35}{3}+40*\frac{3}{3}$

$a_9=\frac{-35}{3}+\frac{120}{3}$

$a_9=\frac{85}{3}$

Hence final answer is n=9, $a_9=\frac{85}{3}$.