Let { a n } be a non - constant arithmetic progression with a 1 = 1 and for any n ≥ 1 , the value a 2 n + a 2 n − 1 + ⋯ + a n + 1 a n + a n − 1 + ⋯ a 1 remains constant. Then, a 15 will be? Options: A . &nbsp 30 B . &nbsp 29 C . &nbsp 31 D . &nbsp Can't be determined

Answer: Option B : B C = a 2 n + a 2 n − 1 + a 2 n − 2 + ⋯ + a n + 1 a n + a n − 1 + ⋯ + a 1 Adding 1 on both sides, we get C + 1 = a 2 n + a 2 n − 1 + a 2 n − 2 + ⋯ + a n + 1 a n + a n − 1 + ⋯ + a 1 = a 2 n + a 2 n − 1 + ⋯ + a n + 1 + a n + ⋯ + a 1 a n + a n − 1 + ⋯ + a 1 ∴ C + 1 = S 2 n S n = 2 n 2 [ 2 a 1 + ( 2 n − 1 ) d ] n 2 [ 2 a 1 + ( n − 1 ) d ] = 2 [ 2 a 1 + ( 2 n − 1 ) d ] 2 a 1 + ( n − 1 ) d Since, C + 1 2 is a constant. Let C + 1 2 = R ⇒ 2 + ( 2 n − 1 ) d 2 + ( n − 1 ) d = R [ ∵ a 1 = 1 ] ⇒ 2 + ( 2 n − 1 ) d = 2 R + ( n − 1 ) d R ⇒ 2 R + n d R − d R = 2 + 2 n d − d ⇒ n d ( R − 2 ) = d R − 2 R − d + 2 ⇒ n d ( R − 2 ) = ( d − 2 ) ( R − 1 ) Now, left side has n which changes and right side remains constant ∴ LHS = RHS = 0 ⇒ R = 2 [ ∵ n ≠ 0 , d ≠ 0 ] ⇒ 0 = d − 2 ⇒ d = 2 ∴ a 15 = a 1 + ( 15 − 1 ) d = 1 + 14 × 2 = 29

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

Discussion Forum : Sequences And Series

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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Discussion Forum : Sequences And Series

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