Sequences And Series Ques 33
- Let $a_1, a_2, a_3, \ldots, a_{49}$ be in AP such that $\sum_{k=0}^{12} a_{4 k+1}=416$ and $a_9+a_{43}=66$. If $a_1^2+a_2^2+\ldots+a_{17}^2=140 \mathrm{m}$, then $m$ is equal to
(2018 Main)
(a) $66$
(b) $68$
(c) $34$
(d) $33$
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Answer:
Correct Answer: 33.(c)
Solution: (c) We have, $a_1, a_2, a_3, \ldots a_{49}$ are in AP.
$ \sum_{k=0}^{12} a_{4 k+1}=416 \text { and } a_9+a_{43}=66 $
Let $\mathrm{a}_1=\mathrm{a}$ and $\mathrm{d}=$ common difference
$\because \quad a_1+a_5+a_9+\cdots+a_{49}=416 $
$\therefore \quad a+(a+4 d)+(a+8 d)+\ldots(a+48 d)=416 $
$\Rightarrow \quad \frac{13}{2}(2 a+48 d)=416 $
$\Rightarrow \quad a+24 d=32 $ $\quad$ ……..(i)
$\text { Also } \quad a a_9+a_{43}=66 $
$\therefore \quad a+8 d+a+42 d=66 $
$\Rightarrow \quad 2 a+50 d=66 $
$\Rightarrow \quad a+25 d=33$ $\quad$ ……..(ii)
Solving Eqs. (i) and (ii), we get
$ a=8 \text { and } d=1 $
Now, $a_1^2+a_2^2+a_3^2+\cdots+a_{17}^2=140 \mathrm{m}$
$ \begin{aligned} & 8^2+9^2+10^2+\ldots+24^2=140 \mathrm{m} \\ & \Rightarrow \quad \left(1^2+2^2+3^2+\ldots+24^2\right)-\left(1^2+2^2+3^2+\ldots+7^2\right)=140 \mathrm{m} \\ & \Rightarrow \quad \frac{24 \times 25 \times 49}{6}-\frac{7 \times 8 \times 15}{6}=140 \mathrm{m} \\ & \Rightarrow \quad \frac{3 \times 7 \times 8 \times 5}{6}(7 \times 5-1)=140 \mathrm{m} \\ & \Rightarrow \quad 7 \times 4 \times 5 \times 34=140 \mathrm{m} \\ & \Rightarrow \quad 140 \times 34=140 \mathrm{m} \Rightarrow \mathrm{m}=34 \\ \end{aligned} $
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