The 17th term of an AP is 5 more than twice its 8th term.

Solution:

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

$a_{17}=2 a_{8}+5$            (Given)

$\therefore a+16 d=2(a+7 d)+5 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+16 d=2 a+14 d+5$

$\Rightarrow a-2 d=-5 \quad \ldots$ (1)

Also,

$a_{11}=43$          (Given)

$\Rightarrow a+10 d=43 \quad \ldots(2)$

From (1) and (2), we get

$-5+2 d+10 d=43$

$\Rightarrow 12 d=43+5=48$

$\Rightarrow d=4$

Puting d = 4 in (1), we get

$a-2 \times 4=-5$

$\Rightarrow a=-5+8=3$

$\therefore a_{n}=a+(n-1) d$

$=3+(n-1) \times 4$

$=4 n-1$

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