Question

please answer my 12 question

Answer 1

Let there be n terms.

$a=5, T_{n}=45, S_{n}=400\\\\T_{n}=45\\a+(n-1)d=45\\5+(n-1)d=45\\(n-1)d=40\\\\S_{n}=400\\\\\frac{n}{2}.[2a+(n-1)d]=400\\\\\frac{n}{2}.[2(5)+40]=400\\\\\frac{n}{2}.[10+40]=400\\\\\frac{n}{2}.(50)=400\\\\25n=400\\n=16$

∴ There are 16 terms.

We have, (n-1)d=40

(16-1)d=40

15d=40

d=$\frac{8}{3}$

∴ Common difference= $\frac{8}{3}$