Chapter 13, Problem 10CR

To determine

To calculate: The ${33}^{\text{rd}}$ term of the sequence $-3,1,5,9,...$ and the sum of the first 20 terms.

Answer to Problem 10CR

Solution:

For the sequence $-3,1,5,9,...$, the ${33}^{\text{rd}}$ term is $125$, and the sum of the first 20 terms is $700$.

Explanation of Solution

Given information:

The sequence $-3,1,5,9,...$

Formula used:

For the arithmetic sequence $\left\{a_n\right\}$, whose first term is $a_1$ and common differenceis $d$, the $n^{th}$ term is defined by the formula $a_n=a_1+\left(n-1\right)d$, and the sum of first $n^{th}$ term is defined by $S_n=\frac{n}{2}\left(a_1+a_n\right)$.

Calculation:

An arithmetic sequence is a sequence where the difference between two consecutive terms is constant.

Consider the sequence $-3,1,5,9,...$

Here, $1-(-3)=4,5-1=4,9-5=4$

The difference between two consecutive terms is constant.

Therefore, the sequence $-3,1,5,9,...$ is an arithmetic sequence.

The sequence $-3,1,5,9,...$ has initial term $a_1=-3$ and the common difference $d=4$.

Using the formula for arithmetic sequence,

$n^{th}$ term is $a_n=a_1+\left(n-1\right)d$

$a_n=-3+\left(n-1\right)4$

$a_n=4n-7$.

Thus, $n^{th}$ term of the arithmetic sequence is $a_n=4n-7$.

Finding ${33}^{rd}$ term of the arithmetic sequence $\left\{a_n\right\}$ is obtained by plugging $n=33$ in $a_n$.

$a_{33}=4(33)-7=125$

Therefore, the ${33}^{rd}$ term of the sequence is $125$.

Now, to find the sum of first $20$ terms, Plug $n=20$ in $S_n=\frac{n}{2}\left(a_1+a_n\right)$,

$\Rightarrow S_{20}=\frac{20}{2}\left(a_1+a_{20}\right)$.

For $a_{20}$, substitute $n=20$ in $a_n=4n-7$,

$\Rightarrow S_{20}=10\left(-3+\left(4(20)-7\right)\right)$

$\Rightarrow S_{20}=10\left(-3+\left(73\right)\right)$

$\Rightarrow S_{20}=10\left(70\right)$

$\Rightarrow S_{20}=700$.

Thus, the sum of first $20$ term is $700$.

Therefore, for the sequence $-3,1,5,9,...$, ${33}^{\text{rd}}$ term is $125$, and the sum of the first 20 terms is $700$.