Question

Let $\left\{a_n\right\}$ be a non - constant arithmetic progression with $a_1=1$ and for any $n\ge 1$, the value $\frac{a_{2n}+a_{2n-1}+\cdots +a_{n+1}}{a_n+a_{n-1}+\cdots a_1}$ remains constant.

Then, $a_{15}$ will be?

• A. 30
• B. 29
• C. 31
• D. Can't be determined

Answer 1

The correct option is B

29

$C=\frac{a_{2n}+a_{2n-1}+a_{2n-2}+\cdots +a_{n+1}}{a_n+a_{n-1}+\cdots +a_1}$
Adding 1 on both sides, we get
$C+1=\frac{a_{2n}+a_{2n-1}+a_{2n-2}+\cdots +a_{n+1}}{a_n+a_{n-1}+\cdots +a_1}$+1
$=\frac{a_{2n}+a_{2n-1}+\cdots +a_{n+1}+a_n+\cdots +a_1}{a_n+a_{n-1}+\cdots +a_1}\therefore C+1=\frac{S_{2n}}{S_n}=\frac{\frac{2n}{2}[2a_1+(2n-1\left)d\right]}{\frac{n}{2}[2a_1+(n-1\left)d\right]}=\frac{2[2a_1+(2n-1\left)d\right]}{2a_1+(n-1)d}$
Since, $\frac{C+1}{2}$ is a constant.
Let $\frac{C+1}{2}=R$
$\Rightarrow \frac{2+(2n-1)d}{2+(n-1)d}=R[\because a_1=1]\Rightarrow 2+(2n-1)d=2R+(n-1)dR\Rightarrow 2R+ndR-dR=2+2nd-d\Rightarrow nd(R-2)=dR-2R-d+2\Rightarrow nd(R-2)=(d-2)(R-1)$
Now, left side has n which changes and right side remains constant
$\therefore$ LHS = RHS = 0
$\Rightarrow R=2[\because n\ne 0,d\ne 0]\Rightarrow 0=d-2\Rightarrow d=2\therefore a_{15}=a_1+(15-1)d=1+14\times 2=29$