Question

How many multiple of 4 lies between 10 and 250? Also find their sum.

Answer 1

Hint: First of all, find the lowest multiple and highest multiple of 4 between 10 and 250. 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 step-by-step 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 250

\[

   \Rightarrow \dfrac{{250}}{4} = 62\dfrac{2}{4} \\

   \Rightarrow \dfrac{{249}}{4} = 62\dfrac{1}{4} \\

   \Rightarrow \dfrac{{248}}{4} = 62 \\

\]

Therefore, the highest multiple is 248.

So, the series is

12, 16, 20, ……………...,248.

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 \[{a_n} = 248\].

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 {a_n} = a + \left( {n - 1} \right)d \\

   \Rightarrow 248 = 12 + \left( {n - 1} \right)4 \\

   \Rightarrow 248 - 12 = 4n - 4 \\

   \Rightarrow 236 + 4 = 4n \\

   \Rightarrow n = \dfrac{{240}}{4} \\

  \therefore n = 60 \\

\]

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{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \\

   \Rightarrow {S_n} = \dfrac{{60}}{2}\left[ {2 \times 12 + \left( {60 - 1} \right)4} \right] \\

   \Rightarrow {S_n} = 30\left[ {24 + 59 \times 4} \right] \\

   \Rightarrow {S_n} = 30\left[ {260} \right] \\

  \therefore {S_n} = 7800 \\

\]

Note: In a given series of A.P. the last term is given by \[{a_n} = a + \left( {n - 1} \right)d\] where \[a\] is the first term, \[d\] is the common difference, \[{a_n}\] is the last term and \[n\] is the total number of terms. The sum of the terms in a given series of A.P. is given by \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].