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Let a₁, a₂, a₃, ....., a₄₉ be in A.P. such that
$\sum_{k=0}^{12}$

a_{4k+1 = 416 and} + $a_{43}$ = 66 . If $a_{2}^{1}$ + $a_{2}^{2}$ + ...... + $a_{2}^{17}$ = 140 m, then m is equal to :

Option 1)
33

Option 2)
66

Option 3)
68

Option 4)
34

Answers (2)

best_answer

$a_{1}+a_{5}+a_{9}+.........+ a_{49}= 416$

If the common diff = d

$\Rightarrow \frac{13}{2}\left [ 2a +12\times 4d\right ]= 416$

$a+24d =32 \rightarrow \left ( 1 \right )$

Also $a_{9}+ a_{43}= a+ 8d+ a+42d=66$

$\Rightarrow a+25d = 33$

Thus $a= 8 , d= 1$

thus $a_{1}^{2}+a_{2}^{2}+....+a_{17}^{2}$ $Thus \sum_{r=1}^{17} \left ( 7+r \right )^{2} = \sum_{r=1}^{17} 49+14r+r^{2}= 4760 = 140m$

Thus $m= 34$

Sum of n terms of an AP -

$S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]$

and

Sum of n terms of an AP

$S_{n}= \frac{n}{2}\left [ a+l\right ]$

- wherein

$a\rightarrow$ first term

$d\rightarrow$ common difference

$n\rightarrow$ number of terms

Option 1)

33

Option 2)

66

Option 3)

68

Option 4)

34

Posted by

Himanshu

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