Question

The sum of the first $n$ terms is $\frac{3n^2}{2}+\frac{13n}{2}$, then ${25}^{th}$ term is

• A. $75$
• B. $80$
• C. $162$
• D. $144$

Answer 1

Answer: (B)

$80$

Let $a_1,d$ and $S_n$ be first term, common difference and the sum of $n$ terms of an AP.
Since, $S_n=\frac{3n^2}{2}+\frac{13n}{2}$ ...(1)
Now, put $n=1$ in equation (1). we get,
$S_1=\frac{3(1{)}^2}{2}+\frac{13\left(1\right)}{2}=\frac{16}{2}=8$
since, $S_1=a_1=8$
Again, put $n=2$ in equation (1). we get,
$S_2=\frac{3(2{)}^2}{2}+\frac{13\left(2\right)}{2}=19$
Since, $S_2=a_1+a_2=19$
$\Rightarrow 8+a_2=19$
$\Rightarrow a_2=11$
So, common difference $d=a_2-a_1=11-8=3$
$\therefore a_{25}=a_1+24d$
$=8+24\times 3=80$
Hence, option B is correct.