Question

If the sum of the series $24,20,16,....$ is $60$. Find the number of terms in the series.

Answer 1

Step 1. Find the number of terms in the series.

As we know that, an arithmetic progression is a sequence of numbers such that the difference $d$ between each consecutive term is a constant.

The series are, $\mathrm{a},\mathrm{a}+\mathrm{d},\mathrm{a}+2\mathrm{d},\mathrm{a}+3\mathrm{d},...$.

The ${\mathrm{n}}^{\mathrm{th}}$ term is,

${\mathbf{T}}_{\mathbf{n}}\mathbf{=}\mathbf{a}\mathbf{+}\left(n-1\right)\mathbf{\times }\mathbf{d}$

Here, ${\mathrm{T}}_{\mathrm{n}}$ is the ${\mathrm{n}}^{\mathrm{th}}$ term, $\mathrm{a}$ is first term, $\mathrm{n}$ is the number of terms in the sequence and $\mathrm{d}$ is the common difference.

Sum of the first ${\mathrm{n}}^{\mathrm{th}}$ terms is given as,

${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)\times d\right]$

Where, ${\mathrm{S}}_{\mathrm{n}}$ is the sum of a term of A. P. , $\mathrm{a}$ is first form of A. P. , $\mathrm{d}$ is common difference and $\mathrm{n}$ is the number of terms.

Step 2. Using the formula of sum of the first ${\mathrm{n}}^{\mathrm{th}}$ term.

We will find the number of terms by using the formula sum of the first $n^{th}$ terms.

${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\times \mathrm{d}\right]$

Here,

${\mathrm{S}}_{\mathrm{n}}=60$

$\mathrm{a}=24$

$\mathrm{d}=20-24$

$=-4$

So,

$\Rightarrow$ $60=\frac{\mathrm{n}}{2}\left[2\left(24\right)+\left(\mathrm{n}-1\right)\times \left(-4\right)\right]$

$\Rightarrow$ $120=\mathrm{n}\left[48-4\mathrm{n}+4\right]$

$\Rightarrow$ $4{\mathrm{n}}^2-52\mathrm{n}+120=0$

Divide each side by $4$.

$\Rightarrow$ ${\mathrm{n}}^2-13\mathrm{n}+30=0$

$\Rightarrow$ $\left(\mathrm{n}-10\right)\left(\mathrm{n}-3\right)=0$

So,

$\Rightarrow$ $\mathrm{n}-10=0$

$\Rightarrow$ $\mathrm{n}=10$

And

$\Rightarrow$ $\mathrm{n}-3=0$

$\Rightarrow$ $\mathrm{n}=3$

Hence, the number of terms in the series is $10$ or $3$.