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Question

Let the sum of the first three terms of an A.P. be 39 and the sum of its last four terms be 178. If the first term of this A.P. is 10, then the median of the A.P. is:

A
29.5
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B
28
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C
31
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D
26.5
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Solution

The correct option is C 29.5Let the first three terms of the A.P. be a−d,a,a+dGiven a−d=10...(1)Also given a−d+a+a+d=39⇒3a=39...(2)Solving (1) and (2), we geta=13,d=3So, the A.P. is 10,13,16,...Now, let the last four terms be an,an−1,an−2,an−3a+(n−1)d+a+(n−2)d+a+(n−3)d+a+(n−4)d=178⇒4a+(4n−10)d=178⇒n=14So, total number of terms in the A.P. are 14 (even).Hence, median of the A.P. will be mean of T7,T8T7=10+6(3)=28T8=10+7(3)=31Median =28+312=29.5

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