Question

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.​

Answer 1

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||✪✪ QUESTION ✪✪||

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

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|| ✰✰ ANSWER ✰✰ ||

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, 115..........995

Here,

• a (first term) = 105
• d (common difference) = 5

Now $\boxed{\bf{\red{a\:n\:=\:a\:+\:(n-1)\:d}}}$

→ 995 = 105 + (n - 5) × 5

→ 890 = (n - 1) × 5

→ 178 = (n - 1)

→ $\boxed{\bf{\purple{n\:=\:179}}}$

$\boxed{\bf{\blue{Sn\:=\:\frac{n}{2}[2a+(n-1)d]}}}$

→ S179 = 179/2 [2 × 105 + (179 - 1) × 5]

→ S179 = (179) [550]

→ $\huge\pink{98450}$

Hence, the sum of all natural numbers lying between 100 and 1000, which are multiplies of 5, is $\red{98450}$

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Answer 2

AnswEr :

$\bf{\pink{\underline{\underline{\bf{Given\::}}}}}$

Numbers lying between 100 and 1000, which are multiples of 5.

$\bf{\red{\underline{\underline{\bf{To\;find\::}}}}}$

The sum of all natural numbers.

$\bf{\orange{\underline{\underline{\bf{Explanation\::}}}}}$

Multiples of 5 between 100 and 1000 are 105, 110, 115, 120,........990, 995.

$\bf{We\:have}\begin{cases}\sf{First\:term\:(a)=105}\\ \sf{Common\:difference=a_{2}-a_{1}=110-105=5}\\ \sf{Last\:term\:(a_{n})=995}\end{cases}}$

A/q

Formula use :

$\bf{\purple{\sf{a_{n}=a+(n-1)d}}}$

→ 995 = 105 + (n-1)5

→ 995 = 105 + 5n - 5

→ 995 = 5n + 100

→ 5n = 995 - 100

→ 5n = 895

→ n = 895/5

→ n = 179

Formula use : (Sum finding)

$\bf{\purple{\sf{S_{n}=\dfrac{n}{2}\ib[ 2a+(n-1)d\big]}}}}}$

→ Sn = 179/2 [2(105) + (179 - 1)5]

→ Sn = 179/2 [210 + 178 × 5]

→ Sn = 179/2 [210 + 890]

→ Sn = 179/2 [1100]

→ Sn = 179/2 × 1100

→ Sn = 179 × 550

→ Sn = 98450

∴ The sum of all naturals number is Sn = 98450 .