Question

Find the sum of the first $25$ terms of an AP, whose $n^{\mathrm{th}}$ term is given by $a_n=7-3n$.

Answer 1

Step 1: Find the first term and common difference:

Substitute $n=1$ in the given $n^{\mathrm{th}}$ term.

$\begin{array}{rcl}& \Rightarrow & a_1=7-3\left(1\right)\\ & =& 7-3\\ & =& 4\end{array}$

Substitute $n=2$ in the given $n^{\mathrm{th}}$ term.

$\begin{array}{rcl}& \Rightarrow & a_2=7-3\left(2\right)\\ & =& 7-6\\ & =& 1\end{array}$

Evaluate $d$ by subtracting $a_1$ from $a_2$ .

$\begin{array}{rcl}& \Rightarrow & d=a_2-a_1\\ & =& 1-4\\ & =& -3\end{array}$

Thus, $a=4$ and $d=-3$.

Step 2: Compute the sum up to $\mathbf{25}$ terms:

Formula:

The sum of $n$ terms of a series is $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.

$\begin{array}{rcl}& \Rightarrow & S_{25}=\frac{25}{2}\left[2\left(4\right)+\left(24\right)\left(-3\right)\right]\\ & =& \frac{25}{2}\left(8-72\right)\\ & =& \frac{25}{2}\left(-64\right)\\ & =& 25\left(-32\right)\\ & =& -800\end{array}$

Therefore, the required sum is $S_{25}=-800$.