Let {an} be a non - constant arithmetic progression with a1=1 and for any n"

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Question

Let

{a_{n}}

be a non - constant arithmetic progression with

a_{1} = 1

and for any

n \geq 1

, the value

\frac{a_{2 n} + a_{2 n - 1} + \dots + a_{n + 1}}{a_{n} + a_{n - 1} + \dots a_{1}}

remains constant.
Then,

a_{15}

will be?

Options:

A . 30

B . 29

C . 31

D . Can't be determined

Answer: Option B
:
B

C = \frac{a_{2 n} + a_{2 n - 1} + a_{2 n - 2} + \dots + a_{n + 1}}{a_{n} + a_{n - 1} + \dots + a_{1}}

Adding 1 on both sides, we get

C + 1 = \frac{a_{2 n} + a_{2 n - 1} + a_{2 n - 2} + \dots + a_{n + 1}}{a_{n} + a_{n - 1} + \dots + a_{1}}

= \frac{a_{2 n} + a_{2 n - 1} + \dots + a_{n + 1} + a_{n} + \dots + a_{1}}{a_{n} + a_{n - 1} + \dots + a_{1}} ∴ C + 1 = \frac{S_{2 n}}{S_{n}} = \frac{\frac{2 n}{2} [2 a_{1} + (2 n - 1) d]}{\frac{n}{2} [2 a_{1} + (n - 1) d]} = \frac{2 [2 a_{1} + (2 n - 1) d]}{2 a_{1} + (n - 1) d}

Since,

\frac{C + 1}{2}

is a constant.
Let

\frac{C + 1}{2} = R

\Rightarrow \frac{2 + (2 n - 1) d}{2 + (n - 1) d} = R [∵ a_{1} = 1] \Rightarrow 2 + (2 n - 1) d = 2 R + (n - 1) d R \Rightarrow 2 R + n d R - d R = 2 + 2 n d - d \Rightarrow n d (R - 2) = d R - 2 R - d + 2 \Rightarrow n d (R - 2) = (d - 2) (R - 1)

Now, left side has n which changes and right side remains constant

∴

LHS = RHS = 0

\Rightarrow R = 2 [∵ n \neq 0, d \neq 0] \Rightarrow 0 = d - 2 \Rightarrow d = 2 ∴ a_{15} = a_{1} + (15 - 1) d = 1 + 14 \times 2 = 29

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