Question

5. Find the sum of the following series

(ii) $182+179+176+....+2$

Answer 1

Step 1: Note the given data

Given series $182+179+176+....+2$

First-term $t_1=182$

Second-term $t_2=179$

Third-term $t_3=176$

Step 2: Find the common difference of A.P.

The general formula of common difference $d={(n+1)}^{th}term-n^{th}term$

Common difference $179-182$

$=-3$

Similarly, $d=t_2-t_1=t_3-t_2=-3$

Given series is an Arithmetic series

Step 3: Finding $n^{th}$ term of the AP in term of $n$

The general formula of $n^{th}$ term of A.P. is $t_n=t_1+(n-1)d$

$$\begin{gathered} =182+(n-1)\times (-3) \\ =182-3n+3 \\ =185-3n \end{gathered}$$

Step 4: Comparing the $n^{th}$ term with given data

Let $t_n=2$

Comparing and we get

$185-3n=2$

$$\begin{gathered} \Rightarrow 3n=185-2 \\ \Rightarrow 3n=183 \end{gathered}$$

$\Rightarrow n=\frac{183}{3}$

$\Rightarrow n=61$

Step 5: Find the sum of the given series

The general formula of the sum of the series $=\frac{n}{2}(t_1+t_n)$

The sum is

$$\begin{gathered} =\frac{61}{2}(182+2) \\ =\frac{61}{2}\times 184 \\ =61\times 92 \\ =5612 \end{gathered}$$

Hence, the sum of the given series $182+179+176+....+2$ is $5612$.