Question

Plz its urgent solve this find the sum of two digit number divisible by 2 or 3.plz fsst

Answer 1

Answer:

3285

Step-by-step explanation:

(1)

Two digit numbers divisible by 2:

Series is 10,12,14...98

First term is 10.

Common difference is 2.

Last term an = 98.

nth term of an AP an = a + (n - 1) * d

98 = 10 + (n - 1) * 2

98 = 10 + 2n - 2

98 = 8 - 2n

90 = 2n

n = 45

Sum of first n terms = n/2[2a + (n - 1) * d]

                                 = 45/2[20 + 44 * 2]

                                 = 45/2[108]

                                 = 2430.

(2)

Two digit numbers divisible by 3:

12,15,18... 99

First term a = 12

Common difference d = 3

Last term an = 99.

nth term of AP = a + (n - 1) * d

99 = 12 + (n - 1) * 3

99 - 12 = 3n - 3

90 = 3n

n = 30

Sum of first n terms = n/2[2a + (n - 1) * d]

                                 = 30/2[24 + (30 - 1) * 3]

                                 = 1665

(3)

Two digit numbers divisible by 6:

12,18,24,..96

First term a = 12

Common difference d = 6

Last term an = 96.

nth term of an AP an = a + (n - 1) * d

96 = 12 + (n - 1) * 6

96 - 12 = 6n - 6

90 = 6n

n = 15

Sum of first n terms = n/2[2a + (n - 1) * d]

                                 = 15/2[24 + 84]

                                 = 810.

So, Sum of two digit numbers = 2430 + 1665 - 810

= 3285

Hence, Sum of two digit numbers divisible by 2 or 3 = 3285.

Hope it helps you

#Berainly