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Question

Question 8
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

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Solution

Given that,a3=12a50=106We know that, an=a+(n1)da3=a+(31)d12=a+2d(i)Similarly, a50=a+(501)d106=a+49d(ii)On subtracting (i) from (ii), we get,94=47dd=2From equation (i), we get,12=a+2(2)a=124=8a29=a+(291)da29=8+(28)2a29=8+56=64Therefore, 29th term is 64.


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