Question

Find the sum of integers between 0 and 500 which are divisible by 7 
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Answer 1

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☆ 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.

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▪︎ 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.
• $\sf a_n = n^{th} \:\:term.$
• d = Common Difference.
• a = First Term.
• $\sf S_n$ = Sum of nth terms.

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▪︎ So, Here:-

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

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

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

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

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

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

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