Question

The sum of the first term and the fifth term of an ascending AP is 26 and the product of the second term by the fourth term is 160. Find the sum of the first seven terms of this AP.

Answer 1

Step 1: Find the value of $\mathit{d}$:

Formulae:

The $n^{\mathrm{th}}$ term of the AP is $a_n=a+\left(n-1\right)d$ and the sum of $n$ terms of an AP is $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$.

The sum of the first term and the fifth term of an ascending AP is $26$.

$\begin{array}{cc}\Rightarrow & a_1+a_5=26\\ \Rightarrow & a+a+\left(5-1\right)d=26\\ \Rightarrow & 2a+4d=26\\ \Rightarrow & a+2d=13\\ \Rightarrow & a=13-2d.......\left(1\right)\end{array}$

The product of the second term by the fourth term is $160$.

$$\begin{gathered} \Rightarrow a_2\times a_4=160 \\ \Rightarrow (a+d)\times (a+3d)=160 \end{gathered}$$

Use equation (1) and simplify.

$\begin{array}{cc}\Rightarrow & (13-2d+d)(13-2d+3d)=160\\ \Rightarrow & (13-d)(13+d)=160\\ \Rightarrow & {\left(13\right)}^2-d^2=160\\ \Rightarrow & 169-d^2=160\\ \Rightarrow & d^2=9\\ \Rightarrow & d=\pm 3\end{array}$

Step 2: Evaluate the sum of the first seven terms of the AP:

Case (i): If $d=3$, then we get,
$\begin{array}{rcl}& \Rightarrow & a=13-2\left(3\right)\\ & =& 13-6\\ & =& 7\end{array}$
The sum of first seven terms with $a=7$ and $d=3$ is as follows.

$\begin{array}{rcl}& \Rightarrow & S_7=\frac{7}{2}\left[2\left(7\right)+\left(7-1\right)\left(3\right)\right]\\ & =& 112\end{array}$

Case (ii): If $d=-3$, then we get,
$\begin{array}{rcl}& \Rightarrow & a=13-2\left(-3\right)\\ & =& 13+6\\ & =& 19\end{array}$
The sum of first seven terms with $a=19$ and $d=-3$ is as follows.
$\begin{array}{rcl}& \Rightarrow & S_7=\frac{7}{2}[2\left(19\right)+\left(7-1\right)\left(-3\right)]\\ & =& 70\end{array}$

Hence, the sum of the first seven terms of the AP is $112,70$.