Question

Find the sum of all two digit numbers which are divisible by 3

Answer 1

Heya

The two digit natural numbers divisible by 3 are

12,15......96,99

The series forms an A.P

an = a + (n-1)d

99 = 12 + (n-1) 3

99-12 = (n-1)3

87/3 = n-1

29 = n-1

29+1 = 30 = n

Sum of n numbers in a series is given by,

Sn = n/2(a+an)

= 30/2 (12+99)

= 15×111

= 1665

Sum of all two digit numbers divisible by 3 is 1665

Hope it helps!

Answer 2

Heya !!!





First two digit number is 10 and Last two Digit
Number is 99.



So we will have to find the sum of Numbers which is divisible by 3 between 10 and 99.




AP = 12 , 15 , 18 , ........99



Here,



First term ( A) = 12


Common Difference ( D) = 3




Tn = 99



A + ( N-1) × D = 99


12 + ( N-1) × 3 = 99


12 + 3N -3 = 99



3N = 99-9



3N = 90



N = 90/3 = 30



Sn = N/2 × [ 2A + ( N-1) × D ]


S30 = 30/2 × [ 2 × 12 + ( 30-1) × 3 ]





=> 15 × (24 + 87)


=> 15 × 111


=> 1665.





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