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

Given, Third term, a3 = 16 As we know, a +(3−1)d = 16 a+2d = 16 ………………………………………. (i) It is given that, 7th term exceeds the 5th term by 12. a7 − a5 = 12 [a+(7−1)d]−[a +(5−1)d]= 12 (a+6d)−(a+4d) = 12 2d = 12 d = 6 From equation (i), we get, a+2(6) = 16 a+12 = 16 a = 4 Therefore, A.P. will be4, 10, 16, 22, …