Question

Find the sum of the numbers lying between $ 107 $ and $ 253 $ that is multiple of $ 5 $

Answer 1

Hint: Since the series will be in A.P so from the series by using the formula $ {a_n} = a + \left( {n - 1} \right)d $ we will get the value of $ n $ and then by using the sum formula which is $ {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $ . we will get the sum of the numbers lying between them.
Formula used:
Sum of an A.P is given by,
 $ {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $
Here,
 $ {S_n} $ , is the sum of an A.P
 $ n $ , is the number of terms
 $ d $ , is the common difference
 $ a $ , is the first term
Another formula we will use is for the $ {n^{th}} $ term of an A.P
 $ {a_n} = a + \left( {n - 1} \right)d $
Here,
 $ {a_n} $ , is the $ {n^{th}} $ term of an A.P

Complete step-by-step answer:
First of all we will see the numbers lying between $ 107 $ and $ 253 $ that is multiple of $ 5 $ . So the numbers are $ 110,115,120,125.....250 $
Now we will see the common difference between them.
So, $ {a_2} - {a_1} = 115 - 110 = 5 $
Similarly, $ {a_3} - {a_2} = 120 - 115 = 5 $
Since, $ {a_3} - {a_2} = {a_2} - {a_1} = 5 $ hence $ d $ will be equal to $ 5 $ and $ a $ will be equal to $ 110 $ and the last term, $ {a_n} = 250 $
So on substituting the values, in the formula $ {a_n} = a + \left( {n - 1} \right)d $ , we get
 $ \Rightarrow 250 = 110 + \left( {n - 1} \right)5 $
Now taking the constant term one side and solving the above equation we get the equation as
 $ \Rightarrow 140 = \left( {n - 1} \right)5 $
On solving for the value of $ n $ , we get
 $ \Rightarrow n - 1 = 28 $
And it will be equal to
 $ \Rightarrow n = 29 $
So now we will find the sum of an A.P. On substituting the values in the formula $ {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $ , we have the equation as
 $ \Rightarrow {S_{29}} = \dfrac{{29}}{2}\left[ {2 \times 110 + \left( {29 - 1} \right)5} \right] $
And on solving the small braces and multiplying it, we get
 $ \Rightarrow {S_{29}} = \dfrac{{29}}{2}\left[ {220 + 140} \right] $
Now on adding and solving the braces, we have
 $ \Rightarrow {S_{29}} = \dfrac{{29}}{2} \times 360 $
On dividing it, we get
 $ \Rightarrow {S_{29}} = 29 \times 180 $
And on multiplying the above constant we get the equation as
 $ \Rightarrow {S_{29}} = 5220 $
Therefore, the sum of the numbers lying between $ 107 $ and $ 253 $ that is multiple of $ 5 $ will be equal to $ 5220 $ .
So, the correct answer is “ $ 5220 $ ”.

Note: In this type of question the only thing we need to know will be the formula, so memorizing the formula is the most important thing we have to do. For calculating the common difference in the A.P we just need to subtract the second term from the first term. So by using this and in this, we solve this type of question.