Let a1,a2,a3,….,a49 be in A.P. such that ∑k=012a4k+1=416 and a9+a43=66 . If a12+a22+….+q172=140m , then m is equal to

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answer is 34.

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Detailed Solution

[SolutionStep1]: Let a1,a2,a3,………,a49 be in A.P Let ' d ' be the common difference [SolutionStep2]: Given ∑k=012a4k+1=416⇒a1+a5+a9+....+a49=416 Here n=13 , common difference D=4d[SolutionStep3]:∴Sn=n22a+n−1D⇒1322a1+124d=416⇒a1+24d=32.................1 Given a9+a43=66⇒a1+8d+a1+42d=66⇒a1+25d=33..............22−1⇒d=1 From (1),a1=8[SolutionStep4]: Let a12+a22+….+a172=S⇒82+92+…..+(24)2=S ∵a17=a1+16d=8+16(1)=24⇒S=12+22+…..+72+82+92+….+(24)2-12+22+….+72=(24)(25)(49)6-7(8)(15)6 ∵∑n2=n(n+1)(2n+1)6=1629400−840=4760=14034∴m=34

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