Question

The eight term of an $A.P.$ is half the second term and eleventh term exceeds one third of its fourth term by $1$. Find its ${15}^{th}$ term

Answer 1

$Letfirsttermbeaandcommondifference=dthena+7d=\left(\frac{1}{2}\right)(a+d)or2a+14d=a+dora=-13dnowa+10d=\left(\frac{1}{3}\right)(a+3d)+1a+10d=\left(\frac{a+3d+3}{3}\right)3a+30d=a+3d+3or3(-13d)+30d=-13d+3d+3-9d+13d-3d=3ord=3soa=-13\times 3=-39\therefore a_{15}=a+14d=-39+14\left(3\right)=-39+42=3$