JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of first \(11\) terms of an \(A.P.\), \(a_{1} a_{2}, a_{3}, \ldots\)is \(0\left(\mathrm{a}_{1} \neq 0\right),\) then the sum of the \(A.P.\), \(a_{1}, a_{3}, a_{5}, \ldots, a_{23}\) is \(k a_{1},\) where \(k\) is equal to
- A \(\frac{121}{10}\)
- B \(-\frac{72}{5}\)
- C \(\frac{72}{5}\)
- D \(-\frac{121}{10}\)
Answer & Solution
Correct Answer
(B) \(-\frac{72}{5}\)
Step-by-step Solution
Detailed explanation
\(a_{1}+a_{2}+a_{3}+\ldots \ldots+a_{11}=0\) \(\Rightarrow\left(a_{1}+a_{11}\right) \times \frac{11}{2}=0\) \(\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{11}=0\) \(\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{1}+10 \mathrm{d}=0\) where d is common difference…
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