Question

the sum of integer from 1 to 100 that are divisible by 2 or 5 is S,then S/50 is​

Answer 1

Step-by-step explanation:

Given:-

The integers from 1 to 100

To find:-

Find the sum of integer from 1 to 100 that are divisible by 2 or 5 is S,then S/50 ?

Solution:-

Integers from 1 to 100 = 1,2,3,...,99,100

List of numbers that are divisible by 2

= 2,4,6,...,98,100

First term = 2

Common difference = 4-2 = 2

Common difference is same throughout the series

They are in the AP

Last term = 100

We know that

nth term of an AP = tn = t1 +(n-1)d

Let tn = 100

=2+(n-1)(2) = 100

=> 2+2n-2 =100

=>2n = 100

=>n = 100/2

=>n = 50

Number of terms that are divisible by 2 = 50

We know that

The sum of first n terms in an AP

=> Sn = (n/2)(t1+tn)

=> S 50 = (50/2)(2+100)

=> S 50 = (25)(102)

=>S 50 = 2550

The sum of the integers from 1 to 100 which are divisible by 2 = 2550 -----(1)

Lst of integers which are divisible by 5 from 1 to 100

=> 5,10,15,20,...,95,100

First term = 5

Common difference = 10-5 = 5

Since the common difference is same throughout the series,they are in the AP

Last term = tn = 100

nth term of an AP = tn = t1 +(n-1)d

Let tn = 100

=5+(n-1)(5) = 100

=> 5+5n-5 =100

=>5n = 100

=>n = 100/5

=>n = 20

Number of terms that are divisible by 5 = 20

The sum of first n terms in an AP

=> Sn = (n/2)(t1+tn)

=> S 20 = (20/2)(5+100)

=> S 20 = (10)(105)

=>S 20 = 1050

The sum of the integers from 1 to 100 which are divisible by 5= 1050 -----(1)

And there are the integers which are divisible by both 2 and 5 i.e. by 10

The list of integers which are divisible by 10

=10,20,...,90,100

Their sum=

=10+20+30+40+50+60+70+80+90+100

=> S10= 550------(3)

So The sum of the integers which are divisible by 2 or 5

=> 2550+1050-550

=>3600-550

=>3050

Given that

The sum of the integers which are divisible by 2 or 5 = S

=> S = 3050

The value of S/50

=> 3050/50

=>61

S/50 = 61

Answer:-

The value of S/50 for the given problem is 61

Used formulae:-

• nth term of an AP = tn = t1 +(n-1)d

• The sum of first n terms in an AP
• => Sn = (n/2)(t1+tn)
• t1 = first term
• n = number of terms
• d = Common difference
• tn = last or nth term