JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(x _{1}, x _{2}, x _{3}, \ldots ., x _{20}\) be in geometric progression with \(x_{1}=3\) and the common ration \(\frac{1}{2}\). A new data is constructed replacing each \(x_{i}\) by \(\left(x_{i}-i\right)^{2}\). If \(\bar{x}\) is the mean of new data, then the greatest integer less than or equal to \(\bar{x}\) is \(.....\)
- A \(143\)
- B \(144\)
- C \(145\)
- D \(142\)
Answer & Solution
Correct Answer
(D) \(142\)
Step-by-step Solution
Detailed explanation
\(\sum x _{0}^{1}=\frac{3\left(1-\left(\frac{1}{2}\right)\right)^{20}}{1 \frac{-1}{2}}=6\left(1-\frac{1}{2^{20}}\right)\) \(=\sum_{ i =1}^{20}\left( x _{ i - i }\right)^{2}\) \(=\sum_{ i =1}^{20}\left( x _{ i }\right)^{2}+( i )^{2}-2 x _{ i } i\) Now…
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