Let a_1,a_2,a_3,,a_49 be in A.P.

Question

Let $a_1,a_2,a_3,\ldots,a_{49}$ be in A.P. such that $\displaystyle\sum_{k=0}^{12}a_{4k+1}=416$ and $a_9+a_{43}=66$. If $a_1^2+a_2^2+\cdots+a_{17}^2=140m$, then $m$ is equal to:

Accepted Answer

a1, a2, a3 . . . a43  are in AP

So, a2 = a1 + d

a3 = a1 + 2d

.

a49 =a1 + 48d

Now given, ${a_9} + {a_{43}} = 66$

$$ \Rightarrow \,\,\,\,$$ a1 + 8d + a1 + 42d = 66

$$ \Rightarrow \,\,\,\,$$ 2a1 + 50d = 66

$$ \Rightarrow \,\,\,\,$$ a1 + 25d = 33 . . . . . (1)

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

$$ \Rightarrow \,\,\,\,$$ a1 + a5 + a9 + a13 +. . . . . 13 items = 416

$$ \Rightarrow \,\,\,\,$$ a1 + a1 +  4d + a1 + 8d + . . . . a1 + 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a1 + 4d +8d + 12d +  . . . . . 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a1 + 4 (1+ 2 + 3 + . . . + 12) d = 416

$$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $$d = 416

$$ \Rightarrow \,\,\,\,$$ 13a1 + 24 $$ \times$$ 13d = 416

$$ \Rightarrow \,\,\,\,$$ a1 + 24 d =32 . . . .(2)

Solving (1) and (2) we get,

d = 1

and  ${a_1} = 8$

$$\therefore\,\,\,$$ a1 = 8

a2 = 8 + 1 = 9

a3 = 8 + 2 = 10

.

a17 = 8 + 16 = 24

Now, $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$$

$$ \Rightarrow \,\,\,\,$$ $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$$

$$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$$

We can write above series like this,

$$ \Rightarrow \,\,\,\,\,$$ (12 +22 + . . . . +242) -  (12 + 22 + . . . . .+ 72) = 140 m

$$ \Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m$$

$$ \Rightarrow \,\,\,\,\,$$ 490 - 140 = 140 m

$$ \Rightarrow \,\,\,\,\,$$4760 = 140 m

$$ \Rightarrow \,\,\,\,\,$$ m = 34

Solutions · 1