Chapter 7.3, Problem 10Q

Solution

To calculate : The rule for $n^{th}$ term $a_n$ of the arithmetic or geometric series, value of $a_{15}$ and the sum of first 15 terms of the sequence $1,7,13,19,\dots$ .

The rule for $n^{th}$ term of the sequence $1,7,13,19,\dots$ is $a_n=6n-5$ , value of $a_{15}$ is 85 and the sum of first 15 terms is 645.

Given information :

The sequence $1,7,13,19,\dots$ .

Formula used :

The $n^{th}$ term of an arithmetic series with first term as $a_1$ and common difference $d$ is given by $a_n=a_1+\left(n-1\right)d$ .

The sum of first $n$ terms of an arithmetic series is $S_n=n\left(\frac{a_1+a_n}{2}\right)$ .

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

The sum of the first $n$ terms of geometric series with first term $a_1$ and common ration $r$ is calculated using the formula $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ with $r\ne 1$ .

Calculation :

Consider the sequence $1,7,13,19,\dots$ ,

It can be observed that each term is 6 more than previous term.

So, the first term is $a_1=1$ and common difference is $d=6$ .

The rule of $n^{th}$ term of an arithmetic series is written as $a_n=a_1+\left(n-1\right)d$ .

  $\begin{array}{c}{a}_n=1+\left(n-1\right)6\\ =1+6n-6\\ a_n=6n-5\end{array}$

Put 15 for $n$ to calculate $a_{15}$ .

  $\begin{array}{c}{a}_{15}=6\left(15\right)-5\\ =90-5\\ =85\end{array}$

The sum of first 15 terms is calculated using the formula $S_n=n\left(\frac{a_1+a_n}{2}\right)$ .

  $\begin{array}{c}{S}_{15}=15\left(\frac{1+85}{2}\right)\\ =15\left(\frac{86}{2}\right)\\ =15\left(43\right)\\ =645\end{array}$

Thus, the rule for $n^{th}$ term of the sequence $1,7,13,19,\dots$ is $a_n=6n-5$ , value of $a_{15}$ is 85 and the sum of first 15 terms is 645.