The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.

Solution:

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

$a_{2}+a_{7}=30$                              (Given)

$\therefore(a+d)+(a+6 d)=30 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 2 a+7 d=30 \quad \ldots$ (1)

Also,

$a_{15}=2 a_{8}-1$              (Given)

$\Rightarrow a+14 d=2(a+7 d)-1$

$\Rightarrow a+14 d=2 a+14 d-1$

$\Rightarrow-a=-1$

$\Rightarrow a=1$

Putting a = 1 in (1), we get

$2 \times 1+7 d=30$

$\Rightarrow 7 d=30-2=28$

$\Rightarrow d=4$

So,

$a_{2}=a+d=1+4=5$

$a_{3}=a+2 d=1+2 \times 4=9$

Hence, the AP is 1, 5, 9, 13,...