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Question

# The sums of first n terms of two A.Pâ€²s are in the ratio (3n+8):(7n+15). The ratio of their 12th terms is

A
49
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B
716
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C
37
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D
167
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Solution

## The correct option is B 716Thereare2ApwithdifferentfirsttermandcommondifferenceforthefirstAPletfirsttermbeacommondifferencebedsumofnterm=Sn=n2[2a+(n−1)d]nthterm=an=a+(n−1)dsimialrlyforsecondtermbe=Acommondifference=DSn=n2[2A+(n−1)]Dnthterm=A+(n−1)Da12offirstAPA12of2ndAP=a+(12−1)dA+(12−1)D=a+11dA+11Dgiventhatsumofntermsof1stAPsumofntermof2ndAP=3n+87n+15n2[2a+(n−1)d]n2[2A+(n−1)D]=3n+87n+152(a+(n−12)d)2(A+(n−12)D)=3n+87n+15a+(n−12)dA+(n−12)D=3n+87n+15→(i)weneedtofinda+11dA+11Dn−12=11n−1=22n=23puttingnineq.(i)a+(23−12)dA+(23−12)D=3×23+87×23+15a+11dA+11D=77176a+11dA+11d=716Theratioof12thtermis716

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