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 ${15}^{th}$ term.

Answer 1

Let ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{d}^{\prime }$ be the first term and common difference of an AP, respectively.

Now, by given condition, $a_8=\frac{1}{2}{a}_2$

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

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

$\Rightarrow a+13d=0$

$\Rightarrow a=-13d\dots \dots \left(i\right)$

And $a_{11}=\frac{1}{3}{a}_4+1$ (given)

$\Rightarrow a+10d=\frac{1}{3}(a+3d)+1$

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

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

$\Rightarrow 2a+27d=3\dots \dots \left(ii\right)$

on Solving the equations gives

$2(-13d)+27d=3$ [Using $eq.\left(i\right)$]

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

$\Rightarrow d=3$

From Eq. (i), we have

$a+13\left(3\right)=0$

$\Rightarrow a=-39$

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