Question

The 17th term of an AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, find its nth term. [CBSE 2012]

Answer 1

Let a be the first term and d be the common difference of the AP. Then,

$a_{17}=2a_8+5$ (Given)
$$\begin{gathered} \therefore a+16d=2\left(a+7d\right)+5\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+16d=2a+14d+5 \\ \Rightarrow a-2d=-5.....\left(1\right) \end{gathered}$$

Also,
$a_{11}=43$ (Given)
$\Rightarrow a+10d=43.....\left(2\right)$

From (1) and (2), we get
$$\begin{gathered} -5+2d+10d=43 \\ \Rightarrow 12d=43+5=48 \\ \Rightarrow d=4 \end{gathered}$$

Puting d = 4 in (1), we get
$$\begin{gathered} a-2\times 4=-5 \\ \Rightarrow a=-5+8=3 \end{gathered}$$

$$\begin{gathered} \therefore a_n=a+\left(n-1\right)d \\ =3+\left(n-1\right)\times 4 \\ =4n-1 \end{gathered}$$

Hence, the n^th term of the AP is (4n − 1).