Question

Find the sum of all natural numbers lying between 100 to 1000 which are multiple of 5.

Answer 1

Hey !!!

________

The first number which is multiple of 5 is 105

and last is 995

it forms an AP with common difference 5 and

995=105+(n-1)5

890=(n-1)5

n=890/5+1 =179

so sum=n/2(a+l)

=179/2(105+995)

=(179×1100)/2=98450


______________

Thanks !!!

Answer 2

$\red\bigstar$QUESTION:-

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

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$\green\bigstar$SOLUTION:-

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Let a= first term of Arithematic progression

⠀⠀ d=common difference of Arithematic progression

⠀⠀ an=nth term of Arithematic progression

⠀⠀ Sn=Sum of nth term of Arithematic progression

All the numbers which are divisible by 5 are listed below

⠀⠀⠀105,110,115,.........,995

Where

•a=105 ,d=110-105=5 and an=995

⠀⠀⠀

Using below formula for finding value of n

$\underline{ \boxed{ \sf \blue{\bigstar \: a_{n} =a + (n - 1)d }}}$

We get ,

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$\longrightarrow \sf \: 995 = 105 + (n - 1)5$

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$\implies \sf \: 995 - 105 = 5n - 5$

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$\sf \implies \: 890 + 5 = 5n$

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$\implies \sf \: n = \dfrac{ \cancel{895} \:}{ \cancel{5}}$

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$\sf \implies \orange{n = 179}$

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Now using below formula for finding sum of nth term

$\underline{ \boxed{ \sf \red{\bigstar \: S_n= \frac{n}{2} \{2a + (n - 1)d \}}}}$

We have

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$\sf \: \: S_n = \dfrac{179}{2} \{2 \times 105 + (179-1)5 \}$

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$\sf \: \: S_n = \dfrac{179}{2} \{210+ 178 \times 5 \}$

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$\sf \: \: S_n = \dfrac{179}{2} \{210 + 891 \}$

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$\sf \: \: S_n = \dfrac{179}{ \cancel{2}} \times \cancel{ 1100}$

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$\sf \: \: S_n = 179 \times 550$

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$\sf \green{ \bigstar \: \: S_n = 98450}$

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Hence ,the sum of all naturals number between 100 to 1000 which is divisible by 5 is 98450