JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(x _1, x _2 \ldots ., x _{100}\) be in an arithmetic progression, with \(x _1=2\) and their mean equal to \(200\) . If \(y_i=i\left(x_i-i\right), 1 \leq i \leq 100\), then the mean of \(y _1, y _2\), \(y _{100}\) is
- A \(10101.50\)
- B \(10051.50\)
- C \(10049.50\)
- D \(10100\)
Answer & Solution
Correct Answer
(C) \(10049.50\)
Step-by-step Solution
Detailed explanation
\(\text { Mean }=200\) \(\Rightarrow \frac{\frac{100}{2}(2 \times 2+99 d)}{100}=200\) \(\Rightarrow 4+99 d =400\) \(\Rightarrow d=4\) \(y_i=i(x i-i)\) \(=i(2+(i-1) 4-i)=3 i^2-2 i\) \(\text { Mean }=\frac{\sum y_i}{100}\) \(=\frac{1}{100} \sum \limits_{i=1}^{100} 3 i^2-2 i\)…
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