Question

find the sum of all natural numbers between 100 and 200 which are divisible by 4
by step by step​

Answer 1

$\huge{\underline{\underline{\bf{Concept}}}}$

• In this question, the concept of Sum of n number of terms is used.
• We have to find the sum of all natural numbers between 100 and 200 which are divisible by 4.

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$\huge{\underline{\underline{\bf{Formula\: used}}}}$

• In this question two formulas are used.

$\large{\boxed{\boxed{\rm{a_n = a + (n - )d}}}}$

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$\large{\boxed{\boxed{\rm{S_n = \dfrac{n}{2} [2a + (n - 1)d]}}}}$

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$\huge{\underline{\underline{\bf{Solution}}}}$

• Let us find the A.P.

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★ Smallest number between 100 and 200 which is divisible by 4 is 104.

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★ Largest number between 100 and 200 which divisible by 4 is 196.

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Note:- Numbers between 100 and 200 means that nunbers from 101 to 199.

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Required A.P

→ 104, 108, 112,... 196

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Here,

• First term (a) = 104
• Common difference (d) = 4
• Last term (an) = 196

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• Let us find the number of terms (n).

$\implies\bf{a_n = 196}$

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$\implies\bf{a + (n - 1)d = 196}$

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$\implies\bf{104 + (n - 1)4 = 196}$

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$\implies\bf{4n - 4 = 196 - 104}$

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$\implies\bf{4n - 4 = 92}$

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$\implies\bf{4n = 92 + 4}$

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$\implies\bf{4n = 96}$

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$\implies\bf{n = \dfrac{96}{4}}$

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$\implies\bf{n = 24}$

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• Now, let us find the sum of all natural numbers between 100 and 200 which are divisible by4.

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$\tt\longmapsto{S_{24} = \dfrac{24}{2} [2(104) + (24 - 1)4]}$

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$\tt\longmapsto{S_{24} = 12[208 + 23(4)}$

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$\tt\longmapsto{S_{24} = 12[208 + 92]}$

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$\tt\longmapsto{S_{24} = 12 \times 300}$

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$\tt\longmapsto{\boxed{S_{24} = 3600}}$

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Hence,

• The sum of all natural numbers between 100 and 200 which are divisible by 4 is 3600.

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Answer 2

Answer:

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