Question

Mohan saves \[Rs.1\] on the first day, \[Rs.2\] on the second day, \[Rs.3\] on the third day and so on. What will be his total savings after 40 days?

Answer 1

Hint:
In this question we see that there is a constant increase in the savings and every day the saving increases by \[Rs.1\]. So in this question the savings are increasing in A.P. series therefore, we use the concepts of A.P. series to find the sum of ‘n’ number of terms. The sum of first ‘n’ terms in an A.P. series having first term ’a’ and common difference ‘d’ is given by: \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] .

Complete step by step solution:
We are given that Mohan saved \[Rs.1\] on the first day, \[Rs.2\] on the second day, \[Rs.3\] on the third day and so on; therefore he has a constant rise of \[Rs.1\] in the savings every day. The saving forms an A.P. series because the everyday increase is a constant.
Therefore the A.P. series so formed is: 1, 2, 3…
In this A.P. series
The first term (a) =1
The common difference (d) = 1
According to the question we need to find the savings after 40 days so we need to find the sum of the first 40 terms of the series.
Therefore we take (n) =40
Now the sum of first ‘n’ terms of A.P. series is given by \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] ,so we put the values a=1, n=40, d=1 in the formula.
We get,
\[
{S_{40}} = \dfrac{{40}}{2}\left[ {2\left( 1 \right) + \left( {40 - 1} \right)1} \right]\\
 = 20\left( {2 + 39} \right)\\
 = 820
\]

So, after 40 days Mohan will save \[Rs.\,820\].

$ \therefore $ Total saving after 40 days is \[Rs.\,820\].

Note:
Just like we can find the sum of first ‘n’ terms of an A.P. series by using the common difference between two terms, we can also find the sum of a finite A.P. series if we know the last term. The sum of all the terms of a finite A.P. series with last term ‘l’ is given by:\[{S_n} = 2n\left( {a + l} \right)\], where ‘n’ is the number of terms and ‘a’ is the first term.