Question

Show that $a_1,a_2,\cdots ,a_n\cdots$ form an AP where $a_n$ is defined as below

$a_n=3+4n$ . Also find the sum of the first $15$ terms in each case.

Answer 1

Step-1 $a_n=3+4n$is general term of AP

The $n$th term is $a_n=3+4n$.

Substitute $n=1,2,3$ into the given $n^{th}$ term:

$\begin{array}{rcl}{a}_1& =& 3+4\left(1\right)\\ & \Rightarrow & a_1=7\\ a_2& =& 3+4\left(2\right)\\ & \Rightarrow & a_2=11\\ a_3& =& 3+4\left(3\right)\\ & \Rightarrow & a_3=15\end{array}$

Thus, the series is $7,11,15,\dots$.

Observe that the common difference in the series is $4$.

Step-2 Sum of $15$ terms of an AP:

Now, apply the formula for the sum of $n$ terms, which is given by ${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)d\right]$.

Since the first term is $a=7$ and the common difference is $d=4$, the sum of $15$ terms is as follows:

$\begin{array}{rcl}{S}_{15}& =& \frac{15}{2}\left[2\left(7\right)+\left(15-1\right)4\right]\\ & \Rightarrow & S_{15}=\frac{15}{2}\left[14+\left(14\right)4\right]\\ & \Rightarrow & S_{15}=\frac{15}{2}\left[70\right]\\ & \Rightarrow & S_{15}=525\end{array}$

Hence, the given $a_n=3+4n$ form an AP and the sum of $15$ terms of a given term is $525$.