Question

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

Answer 1

To Find :-

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

Solution :-

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

It clearly forms a sequence in A.P.

Where,

• First term (a) = 105

• Common difference (d) = 5

nth term of A.P is given as ;

$\red\bigstar$ $\sf{a_n = a + (n - 1)d}$

[ Put the values ]

↪995 = 105 + (n - 1)(5)

↪105 + 5n - 5 = 995

↪ 100 + 5n = 995

↪ 5n = 995 - 100

↪ 5n = 885

↪n = 895/5

↪$\green{\bold{n = 179}}$

Sum of nth term is given as ;

$\green\bigstar$ $\sf{S_n = \frac{n}{2} [2a + (n-1)d]}$

[ Put the values ]

↪ $\sf{S_{179} = \frac{179}{2} [2(105) + (179-1)5]}$

↪ $\sf{S_{179} = \frac{179}{2} [210 + 178 \times 5]}$

↪ $\sf{S_{179} = \frac{179}{2} [210 + 890]}$

↪ $\sf{S_{179} = \frac{179}{2} \times[ 1100]}$

↪ $\sf{S_{179} = 179 \times 550}$

↪ $\red{\bold{S_{179} = 98450}}$

Therefore,

The sum of all natural numbers lying between 100 and 1000, which are multiples of 5 is 98450.

Answer 2

________________

$\red{\underline{\underline{\bigstar\mapsto}}}$QUESTION:-

⠀⠀⠀

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

________________

$\purple{\underline{\underline{\bigstar\mapsto}}}$CONCEPT:-

•Here the concept of Arithematic progression for finding sum of nth term has been used .

________________

$\green{\underline{\underline{\bigstar\mapsto}}}$SOLUTION:-

⠀⠀

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

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

We get ,

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

Now using below formula for finding sum of nth term

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

We have

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

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

⠀⠀⠀

$\large\sf \purple{ \underline{\bigstar \: \: S_n = 98450}}$

⠀⠀⠀

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

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