Question

Find the sum of all three digit natural numbers, which are divisible by 7.​

Answer 1

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

※ The smallest and largest number of 3 digits, which are divisible by 7 are 105 and 994 is respectively.

※ So, the sequence of 3 digit numbers which are divisible by 7 are 105, 112, 119, ....., 994.

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★ We know that

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

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$\begin{lgathered}\begin{lgathered}\begin{lgathered}\tt{Here}\begin{cases} \sf{a = 105}\\ \sf{d = 7}\\ \sf{a_n = 994}\end{cases}\end{lgathered} \:\end{lgathered}\end{lgathered}$

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$\tt\longmapsto{a_n = 994}$

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$\tt\longmapsto{a + (n - 1)d = 994}$

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$\tt\longmapsto{105 + (n - 1) \times 7 = 994}$

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$\tt\longmapsto{7n - 7 = 994 - 105}$

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$\tt\longmapsto{7n - 7 = 889}$

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$\tt\longmapsto{7n = 889 + 7}$

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$\tt\longmapsto{7n = 896}$

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$\tt\longmapsto{n = \dfrac{886}{7}}$

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$\tt\longmapsto{n = 128}$

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★ Now, we will find sum of the given A.P. by using the formula

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$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{\bf{\bigstar{S_n = \dfrac{n}{2}[2a + (n - 1)d]{\bigstar}}}}$

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$\begin{lgathered}\begin{lgathered}\begin{lgathered}\tt{Here}\begin{cases} \sf{a = 105}\\ \sf{d = 7}\\ \sf{n = 128}\end{cases}\end{lgathered} \:\end{lgathered}\end{lgathered}$

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$\tt:\implies\: \: \: \: \: \: \: \: {S_{128} = \dfrac{128}{2}[2(105) + (128 - 1) \times 7]}$

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$\tt:\implies\: \: \: \: \: \: \: \: {S_{128} = 64[210 + (127) \times 7]}$

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$\tt:\implies\: \: \: \: \: \: \: \: {S_{128} = 64[210 + 889]}$

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$\tt:\implies\: \: \: \: \: \: \: \: {S_{128} = 64 \times 1099}$

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$\tt:\implies\: \: \: \: \: \: \: \: {S_{128} = 70336}$

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$\large{\underline{\underline{\green{Required\: Sum = 70,336}}}}$

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Answer 2

Smallest 3 digit number divisible by 7, (a) = 105

Largest 3 digit number divisible by 7, (a_n) = 994

Common difference, (d) = 7

Therefore, the sequence of 3 digit numbers which are divisible by 7 are 105, 112, 119, ....., 994.

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$\star\;{\underline{\frak{We\;know\;that,}}}$

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$\star\;{\boxed{\sf{\purple{a_n = a + (n - 1)d }}}}$

$:\implies\sf a_n = 994$

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$:\implies\sf a + (n - 1)d = 994$

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$:\implies\sf 105 + (n - 1) \times 7 = 994$

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$:\implies\sf 7n - 7 = 994 - 105$

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$:\implies\sf 7n - 7 = 889$

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$:\implies\sf 7n = 889 + 7$

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$:\implies\sf 7n = 896$

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$:\implies\sf n= \cancel{\dfrac{886}{7}}$

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$:\implies\sf n=128$

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$\begin{lgathered}\star\;{\underline{\frak{For\;finding\;sum,}}}\\ \\\end{lgathered}$

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$\star\;{\boxed{\sf{\pink{S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}</p><p> }}}$

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$:\implies\sf S_{128} = \dfrac{128}{2}\bigg\lgroup2(105) + (128 - 1) \times 7\bigg\rgroup$

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$:\implies\sf S_{128} = \dfrac{128}{2}\bigg\lgroup2(105) + (128 - 1) \times 7\bigg\rgroup$

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$:\implies\sf S_{128} = 64\bigg\lgroup210 + (127) \times 7\bigg\rgroup$

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$:\implies\sf S_{128} = 64\bigg\lgroup210 + 889 \bigg\rgroup$

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$:\implies\sf S_{128} = 64 \times 1099$

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$:\implies\sf S_{128} = 70336$⠀⠀⠀