Question

find the sum of all the natural numbers divisible by 5 between 2 and 101​

Answer 1

Answer:-

Step-by-step explanation:

This forms an A.P. with the first term 102 and common difference equal to 3.

⇒303 = 102 + (n –1) 3

⇒303 - 102 = (n –1) 3

⇒201 = (n –1) 3

⇒ n -1 = 67

⇒ n = 68.

102 + 105 + 108….........+ 303 are in AP.

Sn = (n/2)[ 2a + ( n – 1) d]

Sn = (68/2) [ 204 + ( 68 – 1)3]

Sn = (34)[204 + 201]

Sn = 13770.

The integers from 101 to 304, which are divisible by 5, are 105, 110, 115….........300.

This forms an A.P. with the first term 105 and common difference equal to 5.

⇒300 = 105 + (n –1) 5

⇒300 - 105 = (n –1) 5

⇒195 = (n –1) 5

⇒ n - 1 = 39

⇒ n = 40.

105 + 110 + 115….........+ 300 are in AP.

Sn = (n/2)[ 2a + ( n – 1) d]

Sn = (40/2) [ 210 + ( 40 – 1)5]

Sn = (20)[210 + 195]

Sn = 8100.

The integers from 101 to 304, which are divisible by 3 and 5, are 105, 120, 135….........300.

This forms an A.P. with the first term 105 and common difference equal to 15.

⇒300 = 105 + (n –1) 15

⇒300 - 105 = (n –1) 15

⇒195 = (n –1) 15

⇒ n - 1 = 13

⇒ n = 14.

105 + 120 + 135….........+ 300 are in AP.

Sn = (n/2)[ 2a + ( n – 1) d]

Sn = (14/2) [ 210 + ( 14 – 1)15]

Sn = (7)[210 + 195]

Sn = 2835.

∴Required sum= 13770 + 8100 – 2835 = 19035

Thus, the sum of the integers from 101 to 304, which are divisible by 3 or 5, is 19035.

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