Question

The sum of the first three terms of an AP is $33$. If the product of the first and the third terms exceeds the second term by $29$, then find the AP.

Answer 1

Step 1: Find the first term of the AP:

Let the three terms in AP be $a-d,a,a+d$.

The sum of the first three terms of an AP is $33$, so we get,

$\begin{array}{cc}\Rightarrow & a-d+a+a+d=33\\ & 3a=33\\ & a=\frac{33}{3}\\ & =11\end{array}$

Thus, the first term $a=11$.

Step 2: Find the common difference and compute the AP:

Formula:

$\left(a+b\right)\left(a-b\right)=a^2-b^2$

Since the product of the first and the third terms exceeds the second term by $29$, we get,

$$\begin{gathered} \Rightarrow \left(a-d\right)\left(a+d\right)=a+29 \\ \Rightarrow \left(11-d\right)\left(11+d\right)=11+29[\because a=11] \\ \Rightarrow 121-d^2=40 \\ \Rightarrow d^2=81 \\ \Rightarrow d=\pm 9 \end{gathered}$$

If $d=9$, then the AP is of the form $a-d,a,a+d,\dots$ that is, $2,11,20,\dots$.

If $d=-9$, then the AP is of the form $a-d,a,a+d,\dots$ that is, $20,11,2,\dots$.

Hence, the AP are $2,11,20,...$ and $20,11,2,\dots$.