Question

The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12, find the AP. [CBSE 2013C]

Answer 1

Let the first three terms of the AP be (a − d), a and (a + d). Then,
$$\begin{gathered} \left(a-d\right)+a+\left(a+d\right)=48 \\ \Rightarrow 3a=48 \\ \Rightarrow a=16 \end{gathered}$$

Now,
$$\begin{gathered} \left(a-d\right)\times a=4\left(a+d\right)+12\left(\mathrm{Given}\right) \\ \Rightarrow \left(16-d\right)\times 16=4\left(16+d\right)+12 \\ \Rightarrow 256-16d=64+4d+12 \\ \Rightarrow 16d+4d=256-76 \end{gathered}$$
$$\begin{gathered} \Rightarrow 20d=180 \\ \Rightarrow d=9 \end{gathered}$$

When a = 16 and d = 9,
$$\begin{gathered} a-d=16-9=7 \\ a+d=16+9=25 \end{gathered}$$

Hence, the first three terms of the AP are 7, 16 and 25.