Let a1-a2-a3-A49 be in A-P- such that - k-0 12 a 4k-1-416 and a9-a43-66- If a12-a22-a172-140m- then m is equal to-

The correct option is C 34 Given a_1+a_5+a_9+.....+a_49=416 ⇒ a_1+a_1+4d+a_1+8d+......+a_1+48d=416 ⇒ 13a_1+24d+13=416 ⇒ a_1+24d=32 .... ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

Let a1,a2,a...

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

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68

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34

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33

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Solution

The correct option is C $34$
Given
$a_{1} + a_{5} + a_{9} + . . . . . + a_{49} = 416$
$\Rightarrow a_{1} + a_{1} + 4 d + a_{1} + 8 d + . . . . . . + a_{1} + 48 d = 416$
$\Rightarrow 13 a_{1} + 24 d + 13 = 416$
$\Rightarrow a_{1} + 24 d = 32$ ....(i)
$a_{9} + a_{4} 3 = 66$
$a_{1} + 8 d + a_{1} + 42 d = 66$
$\Rightarrow 2 a_{1} + 50 d = 66$
$∴ a_{1} + 25 d = 33$ ...(ii)
saving (i) & (ii)
$a_{1} = 8, d = 1$
$a_{1}^{2} + a_{1}^{2} + . . . . . . a_{1} 7^{2} = 8^{2} + 9^{2} + . . . . {.24}^{2} = 4760$
Given
$4760 = 140 m$
$∴ m = 34$

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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

, then

m

is equal to

Q. 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.

a_{1}^{2} + a_{2}^{2} + \dots + a_{17}^{2} = 140 m

, then

m

is equal to:

Q. Let

a_{1}, a_{2}, a_{3}, a_{4}

be real numbers such that and

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

. Then the smallest possible value of the expression

(a_{1} - a_{2})^{2} + (a_{2} - a_{2})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

lies in interval .

Q. Let

a_{1}, a_{2}, a_{3}

be the first three terms of an A.P. such that

a_{1} + a_{2} + a_{3} = - 12

and

a_{1} a_{2} a_{3} = 80

. Then the value of

a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

is equal to

Q. Let

a_{1}, a_{2}, a_{3}, a_{4}

be real numbers such that

a_{1} + a_{2} + a_{3} + a_{4} = 0

and

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

. Then the smallest possible value of the expression

(a_{1} - a_{2})^{2} + (a_{2} - a_{3})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

lies in the interval

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Let a1,a2,a3,.....A49 be in A.P, such that ∑12k=0a4k+1=416 and a9+a43=66. If a21+a22+....+a217=140m, then m is equal to:

The correct option is C 34Givena1+a5+a9+.....+a49=416⇒a1+a1+4d+a1+8d+......+a1+48d=416⇒13a1+24d+13=416⇒a1+24d=32....(i)a9+a43=66a1+8d+a1+42d=66⇒2a1+50d=66∴a1+25d=33...(ii)saving (i) & (ii)a1=8,d=1a21+a21+......a172=82+92+.....242=4760Given4760=140m∴m=34

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: