Question

The sum of first $5$ terms of an $AP$ is $45$ and that of its first $15$ terms is $435$. Find the sum of first $n$ terms of this $AP$.

• A. $S_n=10n^2-41n$
• B. $S_n=10n^2-21n$
• C. $S_n=56n^2-21n$
• D. None of these

Answer 1

Answer: (D)

None of these

$S_5=45$

$\to \frac{5}{2}(2a+4d)=45$

$S_{15}=435$

$\to \frac{15}{2}(2a+14d)=435$

Solving we get $d=4$ and $a=1$

So $S_n=\frac{n}{2}(2\times 1+(n-1\left)4\right)=2n^2-n$