Question

The first four terms of an AP are $a,9,3a–b,3a+b$. The $2011$th term of an AP is

• A. $2015$
• B. $4025$
• C. $5030$
• D. $8045$

Answer 1

The correct option is D

$8045$

The explanation for the correct answer.

Step 1: Solve the first term and common differences of an AP $a,9,3a–b,3a+b$

$a,9,3a–b,3a+b$

Common difference $d=9-a$

$9-a=3a-b-9$

$4a-b=18..\left(i\right)$

$3$rd term = ($2$nd term + $4$th term)/$2$

$3a-b=\frac{(9+3a+b)}{2}$

$3a-3b=9$

$a-b=3\dots \left(ii\right)$

Solving $\left(i\right)$ and$\left(ii\right)$

We get, $3a=15$

$a=5$

$$\begin{gathered} \Rightarrow d=9-a=9-5 \\ =4 \end{gathered}$$

Solve 2: Solve for the $2011$th term.

${\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathit{a}\mathbf{+}\left(n-1\right)\mathit{d}$

$2011$th term $=a+2010d$

$=5+2010\left(4\right)$

$=5+8040$

$=8045$

Hence, option(D) is the correct answer.