Question

find the
Sum of
two-digit whole numbers
divisible by 3?​

Answer 1

Answer:

1665

First two digit number divisible by 3 = 12

Last two digit number divisible by 3 = 99

∴ First term, (a)=12

Common difference, (d)=3

Last term, (l)=99

n=?

As we know that,

a

n

=a+(n−1)d

∴99=12+(n−1)3

⇒(n−1)

3

87

⇒n=29+1=30

∴ Sum of n terms of an A.P., when last term is known is given by-

S

n

=

2

n

(a+l)

∴S

30

=

2

30

(12+99)

⇒S

30

=15×111=1665

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

According to the question two digit no. s divisible by 3 are 12,15,18,......99

so for calculating the Sn=n/2[2a+(n-1) d]

by using this formula we have to first calculate the n (that is no. of terms)

An=a+(n-1)d

99=12+(n-1) ×3 [ common difference d=3]

99-12=(n-1) ×3

87/3=n-1

29+1=n

n= 30

Now calculating sum of n term, Sn=n/2

[2a+(n-1) d]

Sn=30/2[12+99]

15×111

1665

so, sum of 2 digit terms divisible by 3 is 1665.