Question

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. [CBSE 2014]

Answer 1

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

$a_2+a_7=30$ (Given)
$$\begin{gathered} \therefore \left(a+d\right)+\left(a+6d\right)=30\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 2a+7d=30.....\left(1\right) \end{gathered}$$

Also,
$a_{15}=2a_8-1$ (Given)
$$\begin{gathered} \Rightarrow a+14d=2\left(a+7d\right)-1 \\ \Rightarrow a+14d=2a+14d-1 \\ \Rightarrow -a=-1 \\ \Rightarrow a=1 \end{gathered}$$

Putting a = 1 in (1), we get
$$\begin{gathered} 2\times 1+7d=30 \\ \Rightarrow 7d=30-2=28 \\ \Rightarrow d=4 \end{gathered}$$

So,
$a_2=a+d=1+4=5$
$a_3=a+2d=1+2\times 4=9$,...

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