Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Answer:

option (d) is the answer.

Answer 2

EXPLANATION.

$\implies S_{k} = \dfrac{1 + 2 + 3 + . . . . .+ k}{k}$

S₁² + S₂² + . . . . . + S₁₀² = 5A/12.

As we know that,

$\implies S_{k} = \dfrac{k(k + 1)}{2k} = \dfrac{k + 1}{2}$

$\implies S_{k}^{2} = \bigg(\dfrac{k + 1}{2} \bigg)^{2} = \dfrac{1}{4} (k + 1)^{2}$

$\implies \dfrac{5A}{12} =\displaystyle \sum_{k = 1}^{10} (S_{k})^{2}$

$\implies \dfrac{1}{4} \displaystyle \sum_{k = 1}^{10} (k + 1)^{2}$

$\implies \dfrac{1}{4} \displaystyle \sum_{k = 1}^{10} = (k^{2} + 2k + 1)$

$\implies \dfrac{1}{4} \bigg[\displaystyle \sum_{k = 1}^{10} k^{2} + \sum_{k = 1}^{10} 2k + \sum_{k = 1}^{10} k \bigg]$

$\implies \dfrac{1}{4} \bigg[ \dfrac{k(k + 1)(2k + 1)}{6} + \dfrac{2k(k + 1)}{2} + k \bigg]$

$\implies \dfrac{1}{4} \bigg[ \dfrac{10(11)(21)}{6} + (10)(11) + 10 \bigg]$

$\implies \dfrac{1}{4} \bigg[ 385 + 110 + 10 \bigg]$

$\implies \dfrac{1}{4} \bigg[505 \bigg]$

$\implies \dfrac{5A}{12} = \dfrac{505}{4}$

$\implies A = 101 \times 2 = 303$

Option [D] is correct answer.