Please answer the following questions though it's in one thread

Dear Student,
24: let the first term of AP be a and the common difference be d.
S2n=3.Sni.e. sum of 2n terms of AP = 3 times of the sum of n terms of AP2n2.2a+(2n-1)d=3.n2.2a+(n-1)d2a+(2n-1)d=32.2a+(n-1)d4a+2(2n-1)d=6a+3(n-1)d6a-4a=2(2n-1)d-3(n-1)d2a=[4n-2-3n+3]d2a=(n+1)d ..................(1)
then the required ratio is:
S3nSn=3n2.[2a+(3n-1)d]n2.[2a+(n-1)d]=31×[(n+1)d+(3n-1)d][(n+1)d+(n-1)d]    [from eq(1)]=31×(n+1+3n-1)d(n+1+n-1)d=31×4n2n=6


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Regards

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