Question

Find the sum of all the three digit natural numbers which are multiples of 7.

Answer 1

Step-by-step explanation:

We Have :-

$3 \: digit \: numbers$

To Find :-

$sum \: of \: 3 \: digit \: numbers \: which \: are \: multiples \: of \: 7$

Solution :-

$first \: three \: digit \: number \: divisible \: by \: 7 \: is \: 105 \\ \\ \\ last \: three \: digit \: number \: divisible \: by \: 7 \: is \: 994 \\ \\ \\ so \: the \: series \: is \\ \\ \\ 105, \: 112, \: 119, \: - - - - - - 994 \\ \\ \\ we \: will \: use \: the \: formula \\ \\ \\ sum \: = \dfrac{n}{2} (first \: term \: + last \: term ) \\ \\ \\ so \: to \: find \: n \: we \: will \: use \: \\ \\ \\ a_{n} = a + (n - 1)d \\ \\ \\ putting \: values \\ \\ \\ 994 = 105 + (n - 1)7 \\ \\ \\ (n - 1)7 = 889 \\ \\ \\ n - 1 = 127 \\ \\ \\ n = 128 \\ \\ \\ sum \: = \dfrac{n}{2} (first \: term \: + last \: term ) \\ \\ \\ putting \: values \\ \\ \\ sum = \dfrac{128}{2} (105 + 994) \\ \\ \\ sum = 64(1099) \\ \\ \\ sum = 70336$

Answer 2

$\mathtt{ \huge{ \fbox{Solution :)}}}$

The AP is 105 , 112 , 119 , .... , 994

Here ,

• First term (a) = 105
• Common difference (d) = 7
• Last term (l) = 994

We know that , the nth term of an AP is given by

$\mathtt{ \large{\fbox{ (a)_{n} = a + (n - 1)d}}}$

Thus ,

994 = 105 + (n - 1)7

889/7 = (n - 1)

127 = n - 1

n = 128

Hence , the 128th term of given AP is 994

We know that , the sum of all terms of an AP from its last term is given by

$\large \mathtt{ \fbox{(S)_{n} = \frac{n}{2} (a + l)} }$

Thus ,

Sum = 128/2 × (105 + 994)

Sum = 64 × 1099

Sum = 70336

Hence , the sum of all three digits numbers which are divisible by 7 is 70336

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