Let a1,a2,a3,…………….a49be in A.P. such that∑k=012 a4k+1=416 and a9+a43′=66 ,a12+a22+………...+a172=140m then m is equal to

A
33
B
66
C
68
D
34

Solution:

$\sum_{k = 0}^{12} a_{4 k + 1} = 416$

$\Rightarrow a_{1} + a_{5} + \dots + a_{49} = 416$ $\Rightarrow \frac{13}{2} (a_{1} + a_{49}) = 416$

As $a_{49} = a_{1} + 48 d$ so we have

$\frac{13}{2} (2 a_{1} + 48 d) = 416 \Rightarrow a_{1} + 24 d = 32$ …………………... (i)

Given, $a_{9} + a_{43} = 66 \Rightarrow a_{1} + 8 d + a_{1} + 42 d = 66$

$\Rightarrow a_{1} + 25 d = 33$ …………………………….…(ii)

The equation (i) and (ii) gives $a_{1} = 8, d = 1$

Now, $\begin{array}{l} a_{1}^{2} + a_{2}^{2} + \dots . . + a_{17}^{2} = 8^{2} + 9^{2} + \dots . + 24^{2} \\ = (1^{2} + 2^{2} + \dots . . + 24^{2}) - (1^{2} + 2^{2} + \dots . + 7^{2}) \\ = \frac{24 \cdot 25 \cdot 49}{6} - \frac{7 \cdot 8 \cdot 15}{6} = 4760 = 34 (140) \end{array}$

$∴ m = 34$

Let $a_{1}, a_{2}, a_{3}, \dots \dots \dots \dots \dots .$ $a_{49}$ be in A.P. such that $\sum_{k = 0}^{12} a_{4 k + 1} = 416 and a_{9} + a_{43}^{'} = 66, a_{1}^{2} + a_{2}^{2} + \dots \dots \dots . . . + a_{17}^{2} = 140 m$ then $m$ is equal to

Solution:

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