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

Step 1: Simplify the given condition for finding the value of $a$ and $d$

The given condition are $a_8=\frac{1}{2}\times a_2$ ; $a_{11}=\frac{1}{3}{a}_4+1$

We know that the nth term of an AP is given by the formula $a_n=a+(n–1)d$

$$\begin{gathered} a_8=\frac{1}{2}{a}_2 \\ 2a_8=a_2 \\ 2(a+7d)=a+d \\ 2a+14d=a+d \\ \therefore a=–13d \end{gathered}$$

$$\begin{gathered} a_{11}=\frac{1}{3}{a}_4+1 \\ 3(a+10d)=a+3d+3 \\ 3a+30d=a+3d+3 \\ 2a+27d=3 \end{gathered}$$

Substituting $a=-13d$in the equation,

$$\begin{gathered} 2(-13d)+27d=3 \\ \therefore d=3 \\ \therefore a=-13\left(3\right)=-39 \end{gathered}$$

Step 2: Find the ${15}^{th}$ of the given A.P.

$$\begin{gathered} a_{15}=a+14d \\ =-39+14\left(3\right) \\ =-39+42 \\ =3 \end{gathered}$$

Hence, the ${15}^{th}$ term is $3.$