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Question

The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term and 30th term is 1:3. Calculate the first and the thirteenth term of A.P.

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Solution

Let a be the first term and d be the common difference of the given A.P. Then,
S6=4262{2a+(61)d}=422a+5d=14 ...(i)

It is given that

a10:a30=1:3

a+9da+29d=13

3a+27d=a+29d

2a2d=0

a=d ...(ii)

putting the value of a in (i), we get

2d+5d=14d=2

a=d=2

a13=a+12d=2+2×12=26
Hence, first term =2 and thirteenth term =26

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