Find the sum of integers between 0 and 500 which are divisible by 7 �

1.	Find the sum of integers between 0 and 500 which are divisible by 7
Answer» $\rule{400}4$ ☆ ANSWER:- ▪︎ To Find:- The sum of integers between 0 and 500 which are divisible by 7. ▪︎ So, • The numbers between 0 and 500 are which are divisible by 7 are:- =》 7, 14, 21,......497. $\rule{400}2$ ▪︎ Formulas Used:- $\sf\leadsto a_n = a+(n-1)d. \\ \\ \sf\leadsto S_n = \frac{n}{2}(a+a_n)$ • Where, n = number of TERMS. $\sf a_n = n^{th} \:\:<klux>TERM</klux>.$ d = COMMON Difference. a = First Term. $\sf S_n$ = Sum of nth terms. $\rule{400}2$ ▪︎ So, Here:- $\sf a_n th$ term is 497. d = 7. a = 7. n = ?? an = 497. $\rule{400}2$ $\tt\mapsto \: a + (n - 1)d = 497 \\ \\ \tt\mapsto 7 + (n - 1)7 = 497 \\ \\ \tt\mapsto\cancel{7} + 7n\cancel{ - 7 }= 497. \\ \\ \tt\mapsto7n = 497. \\ \\ \tt\mapsto n = \cancel \frac{497}{7} \\ \\ \tt\mapsto n = 71.$ $\therefore$ There are 71 terms between 0 and 500 which are divisible by 7. $\rule{400}2$ • So sum of 71 terms between 0 and 500 which are divisible by 7 is:- $\\ \tt\mapsto \: S_n = \frac{n}{2} (2a + (n - 1)d) \\ \\ \tt\mapsto \: S_{71} = \frac{71}{2} (7 + 497) \\ \\ \tt\mapsto \: S_{71} = \frac{71}{\cancel2} (\cancel{504}) \\ \\ \tt\mapsto S_{71} = 71 \times 252 \\ \\ \tt\mapsto \: S_{71} = 17892\\$ $\rule{400}4$ ▪︎ Therefore sum of integers between 0 and 500 which are divisible by 7 is 17892. $\rule{400}4$

Find the sum of integers between 0 and 500 which are divisible by 7

Answer»

$\rule{400}4$

☆ ANSWER:-

▪︎ To Find:-

The sum of integers between 0 and 500 which are divisible by 7.

▪︎ So,

• The numbers between 0 and 500 are which are divisible by 7 are:-

=》 7, 14, 21,......497.

$\rule{400}2$

▪︎ Formulas Used:-

$\sf\leadsto a_n = a+(n-1)d. \\ \\ \sf\leadsto S_n = \frac{n}{2}(a+a_n)$

• Where,

n = number of TERMS.
$\sf a_n = n^{th} \:\:<klux>TERM</klux>.$
d = COMMON Difference.
a = First Term.
$\sf S_n$ = Sum of nth terms.

$\rule{400}2$

▪︎ So, Here:-

$\sf a_n th$ term is 497.
d = 7.
a = 7.
n = ??
an = 497.

$\rule{400}2$

$\tt\mapsto \: a + (n - 1)d = 497 \\ \\ \tt\mapsto 7 + (n - 1)7 = 497 \\ \\ \tt\mapsto\cancel{7} + 7n\cancel{ - 7 }= 497. \\ \\ \tt\mapsto7n = 497. \\ \\ \tt\mapsto n = \cancel \frac{497}{7} \\ \\ \tt\mapsto n = 71.$

$\therefore$ There are 71 terms between 0 and 500 which are divisible by 7.

$\rule{400}2$

• So sum of 71 terms between 0 and 500 which are divisible by 7 is:-

$\\ \tt\mapsto \: S_n = \frac{n}{2} (2a + (n - 1)d) \\ \\ \tt\mapsto \: S_{71} = \frac{71}{2} (7 + 497) \\ \\ \tt\mapsto \: S_{71} = \frac{71}{\cancel2} (\cancel{504}) \\ \\ \tt\mapsto S_{71} = 71 \times 252 \\ \\ \tt\mapsto \: S_{71} = 17892\\$

$\rule{400}4$

▪︎ Therefore sum of integers between 0 and 500 which are divisible by 7 is 17892.

$\rule{400}4$