Question

An arithmetic progression starts with a positive fraction and every alternate term is an integer. If the sum of the first $11$ terms is $33$, then find the fourth term.

Answer 1

Let $a$ be the first term and $d$ be their common difference of the AP.

Then, Sum to $n$ terms of an AP $=\frac{n}{2}(2a+(n-1\left)d\right)$

Given, $S_{11}=33$

$=>\frac{n}{2}(2a+(n-1\left)d\right)=33$

$=>\frac{11}{2}(2a+(11-1\left)d\right)=33$

Solving we get

$a+5d=3$

As $a$ is a fraction, $d$ should be the same fraction, so that adding $a+d$ gives an integer second term.

So, $a+5a=3$

$=>a=\frac{1}{2}=d$

$nth$ term of the AP $T_n=a+(n-1)d$

So, $T_4=\frac{1}{2}+(4-1)\frac{1}{2}=2$