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Question

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


A

68

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B

34

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C

33

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D

66

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Solution

The correct option is B

34


Step 1: Simplify the summation

k=012a4k+1=a1+a5+a9+....a49

a1+a5+a9+....a49=a+(a+4d)+(a+8d)+....(a+48d) ...an=a+(n-1)d

a1+a5+a9+....a49=13a+312d=416

13a+312d=416

a+24d=32 ...(i)

Step 2: Obtain second equation for first term and common difference

Given, a9+a43=66

a9+a43=a+8d+a+42d

2a+50d=66

a+25d=33 ...(ii)

Step 3: Solve the results obtained

Subtracting (i) from (ii) we get

d=1

Re-substituting d=1 in (i) we get

a=8

Step 4 : Solve for m

a12+a22+.......+a172=82+92+.....242

a12+a22+.......+a172=(12+22+32+...+242)-(12+22+32+...+72)

Using the formula of sum of squares of first n natural numbers we get

a12+a22+.......+a172=24×25×496-7×8×156 ....[Sn2=nn+12n+16]

a12+a22+.......+a172=4760=140m

m=4760140=34

Hence option (B) is the correct answer.


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