First let’s find the integers from 1 to 100, which are divisible by 2.

And, they are 2, 4, 6… 100.

Clearly, this forms an A.P. with the first term and common difference both equal to 2.

So, we have

100 = 2 + (n –1) 2

n = 50

Hence, the sum is

\( 2+4+6+...+100\)

= \(\dfrac{50}{2}[2(2)+(50-1)(2)]\)

= \(\dfrac{50}{2}[4+8]\)

= \( (25)(102)\)

= 2550

Now, the integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

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

So, we have

100 = 5 + (n –1) 5

5n = 100

n = 20

Hence, the sum is

5+10+....+100

= \( \dfrac{20}{2}[2(5)+(20-1)5]\)

= \( 10[10+(19)5]\)

= \( 10[10+195]\)

\(= 10\times 105=1050\)

Lastly, the integers which are divisible by both 2 and 5, are 10, 20, … 100.

And this also forms an A.P. with the first term and common difference both equal to 10.

So, we have

100 = 10 + (n –1) (10)

100 = 10n

n = 10

10+20+........+100

\( \dfrac{10}{2}[2(10)+(10-1)(10)]\)

= 5[20 + 90]

= 5(110)=550

Thus, the required sum = 2550 + 1050 – 550 = 3050

Therefore, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Answered by Pragya Singh | 2 years agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).