JEE Mains · Maths · STD 11 - 8. sequence and series
Let \({a_1},{a_2},\;.\;.\;.\;.,{a_{49}}\) be in \(A.P.\) such that \(\mathop \sum \limits_{k = 0}^{12} {a_{4k + 1}} = 416\) and \({a_9} + {a_{43}} = 66\). If \(a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m,\) then \(m = \;\;..\;.\;.\;.\;\)
- A \(68\)
- B \(34\)
- C \(33\)
- D \(66\)
Answer & Solution
Correct Answer
(B) \(34\)
Step-by-step Solution
Detailed explanation
(2) \(\because\) \(\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416 \Rightarrow \frac{{13}}{2}\left[ {2{a_1} + 48d} \right] = 416\) \( \Rightarrow {a_1} + 24d = 32\,\,\,\,\,\,\,\,..........\left( 1 \right)\) Now,…
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