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

# Let ${a}_{1},{a}_{2},{a}_{3},\dots \dots \dots \dots \dots .$${a}_{49}$be in A.P. such that then is equal to

1. A

33

2. B

66

3. C

68

4. D

34

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### Solution:

$\sum _{k=0}^{12} {a}_{4k+1}=416$

$⇒{a}_{1}+{a}_{5}+\dots +{a}_{49}=416$ $⇒\frac{13}{2}\left({a}_{1}+{a}_{49}\right)=416$

As ${a}_{49}={a}_{1}+48d$ so we have

$\frac{13}{2}\left(2{a}_{1}+48d\right)=416⇒{a}_{1}+24d=32$     …………………... (i)

Given, ${a}_{9}+{a}_{43}=66⇒{a}_{1}+8d+{a}_{1}+42d=66$

$⇒{a}_{1}+25d=33$         …………………………….…(ii)

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

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

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