2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

From the given statements, we can write, a3 + a7 = 6 …………………………….(i) And a3 ×a7 = 8 ……………………………..(ii) By the nth term formula, an = a+(n−1)d Third term, a3 = a+(3 -1)d a3 = a + 2d………………………………(iii) And Seventh term, a7= a+(7-1)d a7 = a + 6d ………………………………..(iv) From equation (iii) and (iv), putting in equation(i), we get, a+2d +a+6d = 6 2a+8d = 6 a+4d=3 or a = 3–4d …………………………………(v) Again putting the eq.(iii) and (iv), in eq. (ii), we get, (a+2d)×(a+6d) = 8 Putting the value of a from equation (v), we get, (3–4d +2d)×(3–4d+6d) = 8 (3 –2d)×(3+2d) = 8 32 – 2d2 = 8 9 – 4d2 = 8 4d2 = 1 d = 1/2 or -1/2 Now, by putting both the values of d, we get, a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = 1/2 a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2 We know, the sum of nth term of AP is; Sn = n/2 [2a +(n – 1)d] So, when a = 1 and d=1/2 Then, the sum of first 16 terms are; S16 = 16/2 [2 +(16-1)1/2] = 8(2+15/2) = 76 And when a = 5 and d= -1/2 Then, the sum of first 16 terms are; S16 = 16/2 [2.5+(16-1)(-1/2)] = 8(5/2)=20