Question

If the sum of first $75$ terms of an $A.P.$ is $2625$, then the $38^{th}$ term of the $A.P.$ is

• A. 39
• B. 37
• C. 36
• D. 38
• E. 35

Answer 1

Answer: (E)

Let first term and common difference of an $AP$ be a and $d$, respectively.
Given, $S_{75}=2625$
$\therefore 2625=\frac{75}{2}[2\cdot a+(75-1)d]$
$\Rightarrow 2625\cdot 2=75[2a+74d]$
$\Rightarrow 35\cdot 2=2[a+37d]$
$\Rightarrow a+37d=35$
$\Rightarrow a+(38-1)d=35$
$\therefore T_{38}=35$