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Question

Question 16
Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

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Solution

a3=16a+(31)d=16a+2d=16(i)a7a5=12[a+(71)d][a+(51)d]=12(a+6d)(a+4d)=122d=12d=6From equation (i), we get,a+2(6)=16a+12=16a=4Therefore, required A.P is,
4, 10, 16, 22, …


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