Find three consecutive terms in an AP, whose sum and product are 48 and 3840, respectively.

Correct option is D

​$$\begin{aligned}&\text{Let the three consecutive terms in an arithmetic progression be } a - d,\ a,\ \text{and } a + d. \\[5pt]&\text{Given that their sum is 48:} \\&(a - d) + a + (a + d) = 3a = 48 \Rightarrow a = 16 \\[10pt]&\text{Also given that their product is 3840:} \\&(a - d)(a)(a + d) = a(a^2 - d^2) = 3840 \\[10pt]&\text{Substituting } a = 16: \\&16(256 - d^2) = 3840 \Rightarrow 256 - d^2 = 240 \Rightarrow d^2 = 16 \Rightarrow d = \pm 4 \\[10pt]&\text{So the three terms are:} \\&a - d = 16 - 4 = 12,\quad a = 16,\quad a + d = 16 + 4 = 20 \\[10pt]&\text{Hence, the required three consecutive terms in the arithmetic progression are:} \\&{12,\ 16,\ 20}\end{aligned}$$​​