Question

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

Answer 1

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) \\\$$