Question

find the sum of all 2 digit natural number divisible by 5(from A.P. chapter)

Answer 1

Heya

The two digit natural numbers divisible by 5 are

10,20......90,95

The series forms an A.P

an = a + (n-1)d

95 = 10 + (n-1) 5

95-10 = (n-1)5

85/5 = n-1

17 = n-1

17+1 = 18 = n

Sum of n numbers in a series is given by,

Sn = n/2(a+an)

= 18/2 (10+95)

= 9×105

= 945

Sum of all two digit numbers divisible by 5 is 945

Hope it helps!

Answer 2

Heya !!!

First two Digit Number is 10 and last two Digit Number is 99.

We will have to find the sum of the numbers between 10 and 99 which is divisible by 5.

AP = 10 , 15 , 20.......95

Here,

First term ( A) = 10

Common Difference ( D) = 5

Tn = 95

A + ( N-1) × D = 95

10 + ( N-1) × 5 = 95

10 + 5N -5 = 95

10 + 5N = 95+5

10 + 5N = 100

5N = 100-10

5N = 90

N = 90/5 = 18

Therefore,

Sn = N/2 × [ 2A + ( N-1) × D ]

S18 = 18/2 × [ 2 × 10 + ( 18-1) × 5 ]

=> 9 × ( 20 + 85)

=> 9 × 105

=> 945.

Hence,

Sum of the numbers between 10 and 99 which is divisible by 5 is 945.

HOPE IT WILL HELP YOU....... :-)