Question

Find the sum of the natural number 1 and 101 ,which are divisible by 3

Answer 1

The integers from 101 to 304, which are divisible by 3, are 102, 105, 108….........303.
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.