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Question

# The sum of n terms of two arithmetic progressions are in ratio (3n+8):(7n+15). Find the ratio of their (i)12th terms (ii)15th terms.

A
(i)7:16 (ii)95:218
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B
(i)5:19 (ii)105:213
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C
(i)3:20 (ii)83:219
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D
None of these
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Solution

## The correct option is A (i)7:16 (ii)95:218We know that the sum of n terms of an arithmetic progression (AP) is given by: n2×(2a+(n−1)d)where a= first term of the AP d= common difference of the AP Let,a1,d1= first term and common difference of the first AP b1,d2= first term and common difference of the second AP Hence,(2a1+(n−1)d1)(2b1+(n−1)d2)=(3n+8)(7n+15) ---(1)We need to find out ratios of 12th terms of both APswhich is a1+11d1b1+11d2 --(2). Multiply and divide equation (2) with 2which is 2a1+22d12b1+22d2 --(3)On comparing coefficients of common differences we get n−1=22⇒n=23. I.e., if we put n=23 in equ (1) to get ratios of 12th terms.i) Ratios of 12th terms =(3(23)+8)(7(23)+15) =77176 =711Ratios of 15th terms =a1+14d1b1+14d2 --(3). Divide and multiply 2 with equation 3, which is 2a1+28d12b1+28d2 --(4)On comparing coefficients of common differences we get n−1=28On substituting n=29 in equation (1). We getiI) Ratios of 15th terms =(3(29)+8)(7(29)+15) =95218 Thus option A is correct.

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