First slide

Arithmetic progression

Question

$Let a_{1}, a_{2}, a_{3}, \dots a_{49} be in A.P such that \sum_{k = 0}^{12} a_{4 k + 1} = 416 and a_{9} + a_{43} = 66 . If$ ${a_{1}}^{2} + {a_{2}}^{2} + . . + {a_{17}}^{2} = 140 m then m is$

Moderate

A

68
B

34
C

33
D

66

Solution

$\sum_{k = 0}^{12} a_{4 k + 1} = 416 \Rightarrow \frac{13}{2} [2 a_{1} + 48 d] = 416$
$\begin{array}{l} \Rightarrow a_{1} + 24 d = 32 \to (1) \\ \Rightarrow a_{9} + a_{43} = 66 \Rightarrow 2 a_{1} + 50 d = 66 \to (2) \\ (1) and (2) \Rightarrow a_{1} = 8, d_{1} = 1 \\ \sum_{r = 1}^{17} {a_{r}}^{2} = \sum_{r = 1}^{17} [8 + (r - 1)]^{2} = 140 m \\ \Rightarrow \sum_{r = 1}^{17} (r + 7)^{2} = 140 m \\ \Rightarrow \sum_{r = 1}^{17} (r^{2} + 49 + 14 r) = 140 m \\ \Rightarrow \frac{17.18.35}{6} + \frac{14 (17 \times 18)}{2} + 49 \times 17 = 140 m \\ \Rightarrow m = 34 \end{array}$

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