Let a1,a2,a3,.....A49 be in A.P, such that ∑k=012a4k+1=416 and a9+a43=66. If a12+a22+....+a172=140m, then m is equal to:

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Let {a}_{1},{a}_{2},{a}_{3},.....{A}_{49} be in A.P, such that

Question

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

66

68

34

33

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Michael Discussed

Let

a_{1}, a_{2}, a_{3}, ..... A_{49}

be in A.P, such that

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

=416 and

a_{9} + a_{43} = 66

. If

a_{1}^{2} + a_{2}^{2} + .... + a_{17}^{2} = 140 m

, then

m

is equal to:

8 mins ago

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8 mins ago

a_{1} + a_{5} + a_{9} + a_{13} + ...... + a_{49} = 416

6 (a_{1} + a_{49}) + a_{25} = 416

6 (a_{1} + a_{1} + 48 d) + (a_{1} + 24 d) = 416

12 (a_{1} + 24 d) + (a_{1} + 24 d) = 416

13 (a_{1} + 24 d) = 416

a_{1} + 24 d = 32.... (1)

and

a_{9} + a_{43} = 66

a_{1} + 8 d + a_{1} + 42 d = 66

2 a_{1} + 50 d = 66

a_{1} + 25 d = 33..... (2)

By (1) and (2),we get

d = 1

a_{1} = 8

a_{1}^{2} + a_{2}^{2} + .... + a_{17}^{2} = 140 m

a_{1}^{2} + (a_{1} + 1)^{2} + (a_{1} + 2)^{2} + ..... + (a_{1} + 16)^{2} = 140 m

17 a_{1}^{2} + 2 a_{1} (1 + 2 + 3 + ... + 16) + (1^{2} + 2^{2} + .... + 1 6^{2}) = 140 m

17 \times 64 + 2 \times 8 (\frac{16 \times 17}{2}) + \frac{16 \times 33 \times 17}{6} = 140 m

1088 + 2176 + 1496 = 140 m

140 m = 4760

m = 34

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Question Text	Let $a_{1}, a_{2}, a_{3}, ..... A_{49}$ be in A.P, such that $\sum_{k = 0}^{12} a_{4 k + 1}$ =416 and $a_{9} + a_{43} = 66$ . If $a_{1}^{2} + a_{2}^{2} + .... + a_{17}^{2} = 140 m$ , then $m$ is equal to:
Answer Type	Text solution:1
Upvotes	150