Question

(c) Find the sum of all 2-digits natural numbers divisible by 3.​

Answer 1

Step-by-step explanation:

Given:-

2-digits natural numbers

To find:-

Find the sum of all 2-digits natural numbers divisible by 3 ?

Solution:-

The list of all 2-digits natural numbers

= 10,11,12,13,...,99

The list of all 2-digit natural numbers which are divisible by 3

=> Multiples of 3 in 2- digit numbers

=> 12,15,18,...,99

First term of the list = 12

difference between two consecutive numbers

= 15-12=3

=18-15=3

=99-96 = 3

Common difference = 3

The Common difference is same throughout the series

So , 12,15,18,...99 are in the AP

We have

First term (a) = 12

Common difference (d) = 3

Last term (an)=99

We know that.

The general term or nth term or last term in an AP

is an = a+(n-1)d

=>12+(n-1)(3) = 99

=> 12+3n-3 = 99

=> 9+3n = 99

=> 3n = 99-9

=> 3n = 90

=> n = 90/3

=> n = 30

Number of terms in the AP = 30

Now ,

Sum of all terms in the AP

12+15+18+...+99

We know that

The sum of the first n terms in an AP is

Sn = (n/2)[2a+(n-1) d]

On Substituting these values in the above formula

=> S 30 = (30/2)[2(12)+(30-1)(3)]

=> S 30 = (15)[24+(29)(3)]

=> S 30= (15)[24+87]

=> S 30 = (15)(111)

=> S 30 = 1665

(or)

Sum of the first n terms = Sn = (n/2)(a+an)

=> S 30 = (30/2)(12+99)

=> S 30 = (15)(111)

=> S 30 = 1665

Answer:-

The sum of all 2-digits natural numbers divisible by 3 is 1665

Used formulae:-

• The general term or nth term or last term in an AP is an = a+(n-1)d
• Sum of the first n terms =
• Sn = (n/2)(a+an)
• Sn = (n/2)[2a+(n-1) d]
• a = first term
• d = Common difference
• n = Number of terms