Let a1-a2-a3-a49 be in A-P- such that -k-012a4k-1-416 and a9-a43-66- If a21-a22-a217-140m- then m is equal to-

Given: ∑_k=0^12a_4k+1=416 Let common difference of A.P. be d Then, nbsp;∑_k=0^12(a_1+4kd)=416 ⇒ a_1+24d=32 ⋯(1) and a_9+a_43=66 ⇒ a_1+25d=33 ⋯(2) From (1) ...

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Byju's Answer

Standard XII

Mathematics

Geometric Progression

Let a1,a2,a3,...

Question

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

33

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66

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68

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34

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Solution

The correct option is D $34$
Given: $12 \sum k = 0 a_{4 k + 1} = 416$
Let common difference of $A . P .$ be $d$
Then, $12 \sum k = 0 (a_{1} + 4 k d) = 416$
$\Rightarrow a_{1} + 24 d = 32 \dots (1)$
and $a_{9} + a_{43} = 66$
$\Rightarrow a_{1} + 25 d = 33 \dots (2)$
From $(1)$ and $(2),$ we have
$d = 1, a_{1} = 8$
We know that
$1^{2} + 2^{2} + 3^{2} + \dots + 24^{2} = \frac{24 \times 25 \times 49}{6} = 140 \times 35$
Similarly, $1^{2} + 2^{2} + 3^{2} + \dots + 7^{2} = \frac{7 \times 8 \times 15}{6} = 140$
Thus, $17 \sum r = 1 a_{r}^{2}$
$= 8^{2} + 9^{2} + 10^{2} + \dots + 24^{2}$
$= 140 \times 35 - 140 = 140 \times 34$
$∴ m = 34$

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Similar questions

Q. Let

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

be in

A . P .

such that

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

and

a_{9} + a_{43} = 66

. If

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

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Q. Let

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a_{9} + a_{43} = 66

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

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Let a1,a2,a3,…,a49 be in A.P. such that 12∑k=0a4k+1=416 and a9+a43=66. If a21+a22+…+a217=140m, then m is equal to:

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