Let a_1, a_2, dots, a_{49} be in A.P. such th | vedclass

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

Question

(B) Given that $a_1, a_2, \dots, a_{49}$ are in $A.P.$ with common difference $d$.The sum $\sum_{k=0}^{12} a_{4k+1} = a_1 + a_5 + a_9 + \dots + a_{49}

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(B) Given that $a_1, a_2, \dots, a_{49}$ are in $A.P.$ with common difference $d$.The sum $\sum_{k=0}^{12} a_{4k+1} = a_1 + a_5 + a_9 + \dots + a_{49} = 416$.This is an $A.P.$ with $13$ terms,first term $a_1$,and common difference $4d$.Sum $= \frac{13}{2} [2a_1 + (13-1)4d] = 416 \Rightarrow \frac{13}{2} [2a_1 + 48d] = 416 \Rightarrow 13(a_1 + 24d) = 416 \Rightarrow a_1 + 24d = 32 \dots (1)$.Also,$a_9 + a_{43} = (a_1 + 8d) + (a_1 + 42d) = 66 \Rightarrow 2a_1 + 50d = 66 \Rightarrow a_1 + 25d = 33 \dots (2)$.Subtracting $(1)$ from $(2)$,we get $d = 1$. Substituting $d=1$ in $(1)$,$a_1 + 24 = 32 \Rightarrow a_1 = 8$.Now,$\sum_{r=1}^{17} a_r^2 = \sum_{r=1}^{17} [8 + (r-1)1]^2 = \sum_{r=1}^{17} (r+7)^2 = \sum_{r=1}^{17} (r^2 + 14r + 49) = 140m$.Using summation formulas: $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$.For $n=17$: $\frac{17 	imes 18 	imes 35}{6} + 14 	imes \frac{17 	imes 18}{2} + 49 	imes 17 = 1785 + 2142 + 833 = 4760$.$140m = 4760 \Rightarrow m = \frac{4760}{140} = 34$.

Let $a_1, a_2, \dots, a_{49}$ be in $A.P.$ such that $\sum_{k=0}^{12} a_{4k+1} = 416$ and $a_9 + a_{43} = 66$. If $a_1^2 + a_2^2 + \dots + a_{17}^2 = 140m$,then $m = \dots$

Similar Questions

If $\frac{3 + 5 + 7 + \dots \text{ to } n \text{ terms}}{5 + 8 + 11 + \dots \text{ to } 10 \text{ terms}} = 7$,then the value of $n$ is

How many times does the digit $5$ appear when writing natural numbers from $1$ to $100$?

The number of terms in an $A.P.$ is even. The sum of the odd terms is $24$ and the sum of the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$,then the number of terms in the $A.P.$ is:

If $\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i - 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3 = \frac{k}{21}$,then $k$ equals

The sum of $3$ numbers in geometric progression is $38$ and their product is $1728$. The middle number is

Vedclass Test Series

Exam Paper Generator

Online Exam Module