CoormaBak9

2021-08-14

Find the total number of terms in the given arithmetic sequence.
${a}_{1}=-\frac{29}{6},d=-\frac{1}{3},S=-36$

Aubree Mcintyre

Step 1
The first term of the arithmetic sequence is given as $-\frac{29}{6}$
The common difference of the arithmetic sequence is $\frac{1}{3}$
The sum of the terms is given as -36
Step 2
If the number of terms is n, then the sum of the arithmetic sequence is given as
${S}_{n}=\frac{n}{2}\left[2{a}_{1}+\left(n-1\right)d\right]$
$-36=\frac{n}{2}\left[2\cdot \left(-\frac{29}{6}\right)+\left(n-1\right)\cdot \frac{1}{3}\right]$
$-72=n\left[-\frac{29}{6}+\left(n-1\right)\frac{1}{3}\right]$
$-72=-\frac{29n}{3}+n\left(n-1\right)\frac{1}{3}$
$-72=\frac{-29n+n\left(n-1\right)}{3}$
$-72=\frac{-29n+{n}^{2}-n}{3}$
$-216=-30n+{n}^{2}$
${n}^{2}-30n+216=0$
Using the quadratic equation formula, we get
$n=\frac{-\left(-30\right)±\sqrt{{\left(-30\right)}^{2}-4\left(1\right)\left(216\right)}}{2\cdot 1}$
$=\frac{30±\sqrt{900-864}}{2}$
$=\frac{30±\sqrt{36}}{2}$
$=\frac{30±6}{2}$
$=\frac{30-6}{2}$ and $\frac{30+6}{2}$
$=\frac{24}{2}$ and $\frac{36}{2}$
$=12$ and 18
Therefore, the number of terms is either 12 or 18.

Jeffrey Jordon

Answer is given below (on video)

Eliza Beth13

To find the total number of terms in the given arithmetic sequence, we'll use the formula:
$S=\frac{n}{2}\left(2{a}_{1}+\left(n-1\right)d\right)$
Given:
${a}_{1}=-\frac{29}{6}$,
$d=-\frac{1}{3}$,
$S=-36$
Substituting the given values into the formula, we get:
$-36=\frac{n}{2}\left(2\left(-\frac{29}{6}\right)+\left(n-1\right)\left(-\frac{1}{3}\right)\right)$
Simplifying further:
$-36=\frac{n}{2}\left(-\frac{58}{6}-\frac{n-1}{3}\right)$
$-36=\frac{n}{2}\left(-\frac{58}{6}-\frac{n-1}{3}\right)$
$-36=\frac{n}{2}\left(-\frac{58}{6}-\frac{n-1}{3}\right)$
Multiplying through by 6 to clear the fractions:
$-216=n\left(-\frac{58}{1}-2\left(n-1\right)\right)$
$-216=n\left(-58-2n+2\right)$
$-216=n\left(-2n-56\right)$
$-216=-2{n}^{2}-56n$
Rearranging the equation:
$2{n}^{2}+56n-216=0$
$n=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Substituting the values into the formula, we get:
$n=\frac{-56±\sqrt{{56}^{2}-4\left(2\right)\left(-216\right)}}{2\left(2\right)}$
Simplifying:
$n=\frac{-56±\sqrt{3136+1728}}{4}$
$n=\frac{-56±\sqrt{4864}}{4}$
$n=\frac{-56±16\sqrt{19}}{4}$
$n=\frac{-14±4\sqrt{19}}{1}$
Thus, the total number of terms in the given arithmetic sequence is $\overline{)-14±4\sqrt{19}}$.

Mr Solver

Given: ${a}_{1}=-\frac{29}{6}$, $d=-\frac{1}{3}$, and $S=-36$.
We can substitute the given values into the formula and solve for $n$:
$-36=\frac{n}{2}\left(2\left(-\frac{29}{6}\right)+\left(n-1\right)\left(-\frac{1}{3}\right)\right)$
Simplifying the equation:
$-36=\frac{n}{2}\left(-\frac{58}{6}-\frac{n-1}{3}\right)$
Multiplying both sides by 2 to eliminate the fraction:
$-72=n\left(-\frac{58}{6}-\frac{n-1}{3}\right)$
Expanding and rearranging:
$-72=n\left(-\frac{58}{6}-\frac{n}{3}+\frac{1}{3}\right)$
Simplifying further:
$-72=n\left(-\frac{58}{6}-\frac{n}{3}+\frac{1}{3}\right)$
To solve this quadratic equation, we can multiply both sides by 6 to eliminate the fraction:
$-432=n\left(-58-2n+2\right)$
Expanding and rearranging:
$-432=n\left(-2n-56\right)$
Simplifying further:
$-432=-2{n}^{2}-56n$
Rearranging the equation to bring it to standard quadratic form:
$2{n}^{2}+56n-432=0$
Now, we can solve this quadratic equation using any suitable method (factoring, completing the square, quadratic formula, etc.) to find the values of $n$.

Solution:
$S=\frac{n}{2}\left(2{a}_{1}+\left(n-1\right)d\right)$
where:
- $S$ represents the sum of the series
- $n$ represents the number of terms
- ${a}_{1}$ represents the first term of the series
- $d$ represents the common difference between terms
Given:
${a}_{1}=-\frac{29}{6}$
$d=-\frac{1}{3}$
$S=-36$
Substituting the given values into the formula, we have:
$-36=\frac{n}{2}\left(2\left(-\frac{29}{6}\right)+\left(n-1\right)\left(-\frac{1}{3}\right)\right)$
To simplify the equation, we can multiply through by 2 to eliminate the fraction:
$-72=n\left(-\frac{29}{3}+\left(n-1\right)\left(-\frac{1}{3}\right)\right)$
Next, we can distribute the terms inside the parentheses:
$-72=n\left(-\frac{29}{3}-\frac{n-1}{3}\right)$
Now, we can simplify further:
$-72=n\left(\frac{-29-n+1}{3}\right)$
Combining like terms:
$-72=n\left(\frac{-n-28}{3}\right)$
To eliminate the fraction, we can multiply both sides by 3:
$-216=n\left(-n-28\right)$
Expanding the equation:
$-216=-{n}^{2}-28n$
Rearranging the equation to bring all terms to one side:
${n}^{2}+28n-216=0$
To solve this quadratic equation, we can factorize it or use the quadratic formula. Factoring, we have:
$\left(n+18\right)\left(n-12\right)=0$
Setting each factor equal to zero:
$n+18=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}n-12=0$
Solving for $n$:
$n=-18\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}n=12$
Since the number of terms cannot be negative, we discard the solution $n=-18$. Therefore, the total number of terms in the given arithmetic sequence is $n=12$.

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