Question

Find the sum of all multiples of \[4\] between $10$ and $50$?

Answer 1

Hint: First of all, find the lowest multiple and highest multiple of $4$ between $10$ and $50$. Then find the total number of terms of the obtained series which will be in A.P. as they have the same difference. Next find the sum of the terms which is the required answer.

Complete answer:
We know that multiples of $4$ are $4,8,12,16,....$
For the lowest multiple consider $10$
$ \Rightarrow \dfrac{{10}}{4} = 2\dfrac{2}{4}$
$ \Rightarrow \dfrac{{11}}{4} = 2\dfrac{3}{4}$
$ \Rightarrow \dfrac{{12}}{4} = 3$
Therefore, the lowest multiple is $12$.
For the highest multiple consider $50$
$ \Rightarrow \dfrac{{50}}{4} = 12\dfrac{2}{4}$
$ \Rightarrow \dfrac{{49}}{4} = 12\dfrac{1}{4}$
$ \Rightarrow \dfrac{{48}}{4} = 12$
Therefore, the highest multiple is $48$.
So, the series is
$12,16,20,....,48$.
Since the difference is the same or common, the series is in A.P.
In the series, first term $a = 12$, common difference $d = 4$ and last term ${a_n} = 48$.
Here we need to find the number of terms i.e., $n$.
We know that in series of A.P. the last term is given by
${a_n} = a + \left( {n - 1} \right)d$
$ \Rightarrow 48 = 12 + \left( {n - 1} \right)4$
Subtracting $12$ from both sides, we get,
$ \Rightarrow 48 - 12 = \left( {n - 1} \right)4$
$ \Rightarrow 36 = \left( {n - 1} \right)4$
Dividing both sides by $4$, we get,
$ \Rightarrow 9 = \left( {n - 1} \right)$
$ \Rightarrow n = 10$
We know that the sum of the terms in a given series of A.P. is given by
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
$ \Rightarrow {S_n} = \dfrac{{10}}{2}\left[ {2\left( {12} \right) + \left( {10 - 1} \right)4} \right]$
$ \Rightarrow {S_n} = \dfrac{{10}}{2}\left[ {24 + \left( 9 \right)4} \right]$
$ \Rightarrow {S_n} = 5\left[ {24 + 36} \right]$
$ \Rightarrow {S_n} = 5\left[ {60} \right]$
$ \Rightarrow {S_n} = 300$

Note:
For such questions, where the range within which the difference in numbers is small and the overall number of terms will be small, the sum of all the terms can be found manually by finding the terms and adding them also. It will be a simpler and less complex process but may take some time and can result in unwanted mistakes in addition and multiplication. This is not an option if the range is large or if the number of terms will be large.