Question

Find the number of terms of the following AP. $0.6,0.9,1.2,\dots .$ having sum $69$.

Answer 1

Step 1. Find the quadratic equation by using the formula of sum of first $n^{th}$ terms of an AP:

Here, the AP is $0.6,0.9,1.2,\dots .$

The first term, $a=0.6$ and common difference, $d=0.9-0.6=0.3$

Let $n$ terms of the AP must be taken, then

$\Rightarrow S_n=69$
We know that the sum of first $nth$ terms of an AP is $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$.
$\begin{array}{cc}\Rightarrow & \frac{n}{2}\left[2\times \left(0.6\right)+\left(n-1\right)\times \left(0.3\right)\right]=69\\ \Rightarrow & n\left[1.2+0.3n-0.3\right]=138\\ \Rightarrow & n\left[0.3n-0.9\right]=138\\ \Rightarrow & 0.3n^2-0.9n=139\\ \Rightarrow & 3n^2-9n-1380=0\end{array}$

Step 2. Solve the quadratic equation by factorization method:

$$\begin{gathered} \begin{array}{cc}\Rightarrow & n^2-3n-460=0\\ \Rightarrow & n^2+20n-23n-460=0\\ \Rightarrow & n\left(n+\hspace{0.17em}20\right)-23\left(n+\hspace{0.17em}20\right)=0\\ \Rightarrow & \left(n+20\right)\left(n-23\right)=0\end{array} \\ \begin{array}{cccc}\Rightarrow & \left(n+\hspace{0.17em}20\right)=0& and& \left(n-23\right)=0\\ \Rightarrow & n=-20& and& n=23\end{array} \end{gathered}$$

Step 3. Find the possible number of terms:

$n=-20$ is neglected as $n$ being the number of terms, is a natural number.
$\therefore n=23$

Hence, the sum of $23$ terms of the AP is $69$.