Question

find the sum of the numbers lying between 1 and 100 which are divisible by 2or5.

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Answer 1

Answer:

Step-by-step explanation:

2,4,5,6,8,10,12,14,15,16,18,20,22,24,25,26,28,30,32,34,35,36,38,40,42,44,46,48,50,52,54,55,56,58,60,62,64,65,66,68,70,72,74,75,76,78,80,82,84,85,86,88,90,92,94,95,96,98,100

Answer 2

Answer:

Sum of numbers lying between 1 and 100 which are divisible by 2 or 5 is 68.

Step-by-step explanation:

From the properties of arithmetic progressions :

• $a_n=a+(n-1)d$, where a$_{n}$ is the nth term, a is the first term, n is the number of terms and d is the common difference between the terms.

Here,

First number between 1 and 100 which is divisible by 2 = 2

First number between 1 and 100 which is divisible by 5 = 5

Last number between 1 and 100 which is divisible by 2 = 98

Last number between 1 and 100 which is divisible by 5 = 95

For 2 : Let the number of terms be n.

= > Last term = First term + ( n - 1 ) common difference

= > 98 = 2 + ( n - 1 )2

= > 98 - 2 = ( n - 1 )2

= > 96 = ( n - 1 )2

= > 96 / 2 = n - 1

= > 48 = n - 1

= > 48 + 1 = n

= > 49 = n

For 5 : Let the number of terms be k.

= > Last term = First term + ( k - 1 ) common difference

= > 95 = 5 + ( k - 1 )5

= > 95 - 5 = ( k - 1 )5

= > 90 = ( k - 1 )5

= > 90 / 5 = k - 1

= > 18 = k - 1

= > 18 + 1 = k

= > 19 = n

Hence the sum of numbers lying between 1 and 100 which are divisible by 2 or 5 is n + k i.e. 49 + 19 i.e. 68.