JEE Mains · Maths · STD 11 - 8. sequence and series
Let \( \sum_{k=1}^{n}a_{k}=\alpha n^{2}+\beta n \). If \( a_{10}=59 \) and \( a_{6}=7a_{1} \) then \( \alpha+\beta \) is equal to:
- A 12
- B 3
- C 5
- D 7
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
\(a_n=S_n-S_{n-1}\) \(=\left(\alpha n^2+\beta n\right)-\left(\alpha(n-1)^2+\beta(n-1)\right)\) (1) \(a _{59} \Rightarrow 19 \alpha+\beta=59\) (2) \(a_6=7 a_1 \Rightarrow 11 \alpha+\beta=7(\alpha+\beta)\) \(\Rightarrow 2 \alpha=3 \beta\) \(a=3, \beta=2\) \(\alpha+\beta=5\)
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