Question

Question 3
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 a and d be the first term and common difference of an AP, respectively.
Now, by given condition, $a_8=\frac{1}{2}{a}_2$
$\Rightarrow a+7d=\frac{1}{2}(a+d)[\because a_n=a+(n-1\left)d\right]$
$\Rightarrow 2a+14d=a+d$
$\Rightarrow a+13d=0$ ...(i)
$And{a}_{11}=\frac{1}{3}{a}_4+1$
$\Rightarrow a+10d=\frac{1}{3}(a+3d)+1$
$\Rightarrow 3a+30d=a+3d+3$
$\Rightarrow 2a+27d=3$
Solving the equations gives,
2(-13d) + 27d = 3
$\Rightarrow -26d+27d=3$
$\Rightarrow d=3$
From Eq. (i),
a + 13 (3) = 0
$\Rightarrow$ a = -39
$\therefore a_{15}=a+14d=-39+14\left(3\right)$
= -39 + 42 = 3