Question

how many nutural numbers are there in the first 100 natural numbers which are multiple of both 2&7​

Answer 1

Step-by-step explanation:

ANSWER

The three digit numbers which are multiples of 7 are

105,112,119,...,994

a

2

−a

1

=112−105=7

a

3

−a

2

=119−112=7

∵a

3

−a

2

=a

2

−a

1

=7

Therefore, the series is in AP

Here, a=105,d=7 and a

n

=994

We know that,

a

n

=a+(n−1)d

⇒994=105+(n−1)7

⇒994−105=(n−1)7

⇒889=(n−1)7

⇒127=(n−1)

⇒n=128

Now, we have to find the sum of this AP

S

n

=

2

n

[2a+(n−1)d]

⇒S

128

=

2

128

[2×105+(128−1)7]

⇒S

128

=64[210+127×7]

⇒S

128

=64[1099]

⇒S

128

=70336

Hence, the sum of all three digit numbers which are

multiple of 7 are 70336.

Answer 2

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Natural numbers which are multiple of both 2&7 are LCM of 2 and 7 i.e. 14

So, natural numbers which are multiple of 14 in the first 100 natural numbers are 14, 28,...,98

Its form an AP series, with first term, a = 14, common difference, d = 14, and last term, an = 98.

We know, an = a + (n - 1)× d

98 = 14 + (n - 1) × 14

98 - 14 = (n - 1) × 14

84 = 14n - 14

98 = 14n

n = 7