Question

The sum of 2+4+6…………..+80 is?
(a) 1540
(b) 1640
(c) 1740
(d) 1440

Answer 1

Hint: Notice that the terms of the series given in the question are in arithmetic progression with common difference and the first term equal to 2. So, directly use the formula of sum of an A.P. to get the answer.

Complete step-by-step answer:
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{r}}$, and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)=\dfrac{r}{2}\left( a+l \right)$
Now moving to the series given in the question. We can see that the terms of the series given in the question are in A.P. with common difference equal to 2 and the first term of the A.P. being 2 as well.
So, using the formula for ${{r}^{th}}$ term of an A.P., we get
${{T}_{r}}=a+\left( r-1 \right)d$
$\Rightarrow 80=2+\left( r-1 \right)2$
$\Rightarrow 78=\left( r-1 \right)2$
$\Rightarrow r=40$
Now to find the sum of the A.P., we will use the formula ${{S}_{r}}=\dfrac{r}{2}\left( a+l \right)$ . In the formula a is the first term, so a=2 and l is the last term so l=80. Therefore, the sum of the series 2+4+6…………..+80 is:
${{S}_{r}}=\dfrac{r}{2}\left( a+l \right)$
$\Rightarrow {{S}_{r}}=\dfrac{40}{2}\left( 2+80 \right)$
${{S}_{r}}=40\times 41=1640$
Therefore, we can conclude that the answer to the above question is option (b).

Note: It is not always necessary that you can notice that the sequences are in the form of simple arithmetic progressions and draw results, but it is for sure that the terms of a sequence will have some order, and it depends on you how wisely you figure out the pattern. The suggested approach can be to try to find the general term of the sequence by observing the pattern of numbers appearing in the sequence and proceed.