Let \(a_1, a_2, a_3, \ldots\) be in an arithmetic progression of positive terms. Let \(A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots + a_{2k-1}^2 - a_{2k}^2\). If \(A_3 = -153, A_5 = -435\) and \(a_1^2 + a_2^2 + a_3^2 = 66\), then \(a_{17} - A_7\) is equal to ____.

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Sequences and Series

Let

a_{1}, a_{2}, a_{3}, \dots

be in an arithmetic progression of positive terms. Let

A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}

. If

A_{3} = - 153, A_{5} = - 435

and

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 66

, then

a_{17} - A_{7}

is equal to ____.

Answer: 910

With common difference

d

, pair terms:

a_{2 k - 1}^{2} - a_{2 k}^{2} = (a_{2 k - 1} - a_{2 k}) (a_{2 k - 1} + a_{2 k}) = (- d) (a_{2 k - 1} + a_{2 k})

, so

A_{k} = - d k [2 a_{1} + (2 k - 1) d]

A_{3} = - 153 \Rightarrow d [2 a_{1} + 5 d] = 51

;

A_{5} = - 435 \Rightarrow d [2 a_{1} + 9 d] = 87

. Subtracting:

4 d^{2} = 36 \Rightarrow d = 3

, then

a_{1} = 1

(positive terms). Check

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 1 + 16 + 49 = 66

a_{17} = 1 + 16 (3) = 49

, and

A_{7} = - 3 \cdot 7 [2 + 13 \cdot 3] = - 21 \cdot 41 = - 861

. Thus

a_{17} - A_{7} = 49 - (- 861) = 910.

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Integer answer