**Question:**

The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.

**Solution:**

Let a and d be the first term and common difference of an AP respectively.

Now, by given condition, $a_{8}=\frac{1}{2} a_{2}$

$\Rightarrow$ $a+7 d=\frac{1}{2}(a+d)$ $\left[\because a_{n}=a+(n-1) d\right]$

$\Rightarrow$ $2 a+14 d=a+d$

$\Rightarrow$ $a+13 d=0$ $\ldots($ i)

and $a_{11}=\frac{1}{3} a_{4}+1$

$\Rightarrow$ $a+10 d=\frac{1}{3}[a+3 d]+1$

$\Rightarrow \quad a+10 d=\frac{1}{3}[a+3 d]+1$

$\Rightarrow \quad 3 a+30 d=a+3 d+3$

$\Rightarrow \quad 2 a+27 d=3$ ...(ii)

From Eqs. (i) and (ii),

$2(-13 d)+27 d=3$

$\Rightarrow \quad-26 d+27 d=3$

$\Rightarrow \quad d=3$

From Eq. (i),

$a+13(3)=0$

$\Rightarrow \quad a=-39$

$\therefore$ $a_{15}=a+14 d=-39+14(3)$

$=-39+42=3$