JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(S\) be the sum of the first \(9\) terms of the series: \(\{x+k a\}+\left\{x^{2}+(k+2) a\right\}+\left\{x^{3}+(k+4) a\right\}+\) \(\left\{x^{4}+(k+6) a\right\}+\ldots \ldots\) where \(a \neq 0\) and \(x \neq 1 .\) If \(S =\frac{ x ^{10}- x +45 a ( x -1)}{ x -1},\) then \(k\) is equal to
- A \(-5\)
- B \(1\)
- C \(-3\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(-3\)
Step-by-step Solution
Detailed explanation
\(S =[ x + ka +0]+\left[ x ^{2}+ ka +2 a \right]+\left[ x ^{3}+ ka +\right.\) \(4 a]+\left[x^{4}+k a+6 a\right]+\ldots \ldots 9\) terms \(\Rightarrow S =\left( x + x ^{2}+ x ^{3}+ x ^{4}+\ldots . .9\right.\) terms \()+( ka + ka\) \(+ ka + ka +\ldots \ldots .9\) terms…
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