JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of the first ten terms of the series \({\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + \;\;.\;.\;.\;.\;,\) is \(\frac{{16}}{5}m\) then \(m\) is equal to :
- A \(100\)
- B \(99\)
- C \(102\)
- D \(101\)
Answer & Solution
Correct Answer
(D) \(101\)
Step-by-step Solution
Detailed explanation
\(\left(1 \frac{3}{5}\right)+\left(2 \frac{2}{5}\right)^{2}+\left(3 \frac{1}{5}\right)^{2}+4^{2}+\left(4 \frac{4}{5}\right)^{2}+\ldots \ldots\) upto \(10\) terms…
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