If the sum of the first 11 terms of the serie | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given series is: $\left(\frac{11}{7}\right)^{2} + \left(\frac{12}{7}\right)^{2} + \left(\frac{13}{7}\right)^{2} + \ldots$ The $n^{th}$ term is

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(C) The given series is: $\left(\frac{11}{7}ight)^{2} + \left(\frac{12}{7}ight)^{2} + \left(\frac{13}{7}ight)^{2} + \ldots$ The $n^{th}$ term is given by $\left(\frac{10+n}{7}ight)^{2}$. The sum of the first $11$ terms is $S_{11} = \sum_{n=1}^{11} \left(\frac{10+n}{7}ight)^{2} = \frac{1}{49} \sum_{n=1}^{11} (10+n)^{2}$. Expanding the sum: $S_{11} = \frac{1}{49} [11^2 + 12^2 + 13^2 + \ldots + 21^2]$. Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get: $S_{11} = \frac{1}{49} \left[ \sum_{k=1}^{21} k^2 - \sum_{k=1}^{10} k^2 ight] = \frac{1}{49} \left[ \frac{21 	imes 22 	imes 43}{6} - \frac{10 	imes 11 	imes 21}{6} ight]$. $S_{11} = \frac{1}{49} 	imes \frac{21}{6} [22 	imes 43 - 10 	imes 11] = \frac{1}{49} 	imes \frac{7}{2} [946 - 110] = \frac{1}{14} 	imes 836 = \frac{11}{7} 	imes 38$. Comparing this with $\frac{11}{7}\lambda$,we get $\lambda = 38$.

If the sum of the first $11$ terms of the series ${\left( {1\frac{4}{7}} \right)^2} + {\left( {1\frac{5}{7}} \right)^2} + {\left( {1\frac{6}{7}} \right)^2} + {2^2} + {\left( {2\frac{1}{7}} \right)^2} + ......$ is $\frac{{11}}{7}\lambda $,then $\lambda $ is equal to:

Similar Questions

If $a, b, c$ are in $GP$ and $4a, 5b, 4c$ are in $AP$ such that $a + b + c = 70$,then the value of $a^3 + b^3 + c^3$ is

The sum of $p$ terms of an $A.P.$ is $3p^2 + 4p$. Find the $n^{th}$ term.

The first term of an $A.P.$ of consecutive integers is $p^2 + 1$. The sum of $(2p + 1)$ terms of this series can be expressed as:

The common ratio of a $G.P.$ is $-\frac{4}{5}$ and the sum to infinity is $\frac{80}{9} .$ Find the first term.

If $a, b, c, d, e, f$ are in $A.P.$,then the value of $e - c$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module