Question

Find the sum of all integers between 50 and 500; which are divisible by 7.

Answer 1

Hint:
Arithmetic progression: An arithmetic progression or sequence is the sequence of numbers such that the difference the consecutive terms is finite.
Arithmetic progression will be in \[a + d, a + 2d,...\]form.
Some general formulas to solve such question:
\[{a_n} = a + (n - 1)d\]
\[{S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)\] where \[{a_n}\]=last term; a=first term; d=common difference; n=no of terms and \[{S_n}\] is summation of series upto n terms.

Complete step by step solution:
As we all know that \[56, 63, 70...497\]will be the integers between 50 and 500 that are divisible by 7.
Each integer is having a common difference of 7 will result in an Arithmetic progression series.
The series \[56,63,70...497\]will have a common difference of 7 with first term 56 and last term 497.
Using formula, \[{a_n} = a + (n - 1)d\], we can find the total number of terms i.e. n.
\[
   = {a_n} = a + (n - 1)d \\
   \Rightarrow 497 = 56 + (n - 1)7 \\
   \Rightarrow 441 = (n - 1)7 \\
   \Rightarrow 63 = (n - 1) \\
   \Rightarrow n = 64 \\
\]
Finding summation of series using: \[{S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)\]
\[{S_n}\]=Summation of series upto nth term.
\[
  {S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right) \\
   \Rightarrow {S_{64}} = \dfrac{{64}}{2}\left\{ {2(56) + (64 - 1)7} \right\} \\
\]
\[ \Rightarrow {S_{64}} = 32\left( {112 + 441} \right)\]
\[ \Rightarrow {S_{64}} = 32\left( {553} \right)\]
\[ \Rightarrow {S_{64}} = 17696\]
So, the sum of all integers between 50 and 500; which are divisible by 7 is \[17696\].

Note:
Student’s can use this formula for finding summation of series upto n terms to make calculations easier: \[{S_n} = \dfrac{n}{2}(a + {a_n})\].
\[
  \because {S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right) \\
   \Rightarrow {S_n} = \dfrac{n}{2}\left( {a + a + (n - 1)d} \right) \\
   \Rightarrow {S_n} = \dfrac{n}{2}(a + {a_n})......(\because {a_n} = a + (n - 1)d) \\
\]