Question

17. How many terms of the AP: $9,17,25.....$ must be taken to give a sum of $636$?

Answer 1

Step 1. Note the given data:

Given an A.P. $9,17,25.....$

First term $t_1=9$

Second term $t_2=17$

Third term $t_3=25$

Sum of $n$ terms of an AP is $S_n=636$

Step 2. Find the common difference:

The formula for the common difference of an AP is $d=n^{th}term-{\left(n-1\right)}^{th}term$

The common difference is

$$\begin{gathered} d=t_2-t_1 \\ =17-9 \\ =8 \end{gathered}$$

Similarly $d=t_3-t_2=8$

Step 3. Find the sum of the first $n$ term of the AP:

The formula of the sum of the first $n$ term of an AP is$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Here $a=9,d=8$

The sum of the first $n$ term of the AP is $S_n=\frac{n}{2}\left[2·9+\left(n-1\right)8\right]$

$$\begin{gathered} =\frac{n}{2}\left[18+8n-8\right] \\ =\frac{n}{2}\left[10+8n\right] \\ =\frac{n}{2}·2\left[5+4n\right] \\ =n\left(5+4n\right) \\ =4n^2+5n \end{gathered}$$

Assume that the sum of first $n$ term of the AP is $636$.

Step 4. Equating $S_n$ with $636$:

$636=4n^2+5n$

Subtracting $636$ on both sides,

$$\begin{gathered} 4n^2+5n-636=636-636 \\ 4n^2+5n-636=0 \end{gathered}$$

Step 5. Find the real roots of the quadratic equation by the factorization method:

Splitting method: If $a{x}^2+bx+c=0$ is an equation

(i) Split the coefficient of $x$such that $b=p+q$ and $ac=pq$

$$\begin{gathered} 4n^2+5n-636=0 \\ 4n^2+53n-48n-636=0 \\ n\left(4n+53\right)-12\left(4n+53\right)=0 \end{gathered}$$ $[\sin ce53+\left(-48\right)=5and53·\left(-48\right)=4·\left(-636\right)]$

We can also write as,

$\left(n-12\right)\left(4n+53\right)=0$

Either

$$\begin{gathered} n-12=0 \\ n=12 \end{gathered}$$

Or,

$$\begin{gathered} 4n+53=0 \\ 4n=-53 \\ n=\frac{-53}{4} \end{gathered}$$

The number of terms cannot be negative.

Therefore, $n=12$

Hence, $636$ is the sum of the first $12$ terms of a given A.P.