Question

solve this question ( maths )

Answer 1

The first two digit no. divisible by 3 is 12
a=12
d=3
And the last 2 digit no. is 99
a(n)=99
Therefore,
a(n)=a+(n-1)d
99=12+(n-1)3
99=12+3n-3
99-12+3=3n
90=3n
n=30
Now,
S(n)=n/2[a+a(n)]
= 30/2[12+99]
=15*111
=1665



HOPE IT HELPS..

Answer 2

First 2 digit no. Which is divisible by 2 is 10
Last 2 digit no. which is divisible by 2 is 98
So, 10,12,14,16..........,98
a =10
d = 12-10 = 2
l = 98
a+(n-1)d= 98
10+(n-1)(2) = 98
10+2n-2 = 98
2n+8=98
2n=98-8
2n=90
n=90/2
n=45

Now, Sn = n/2 [2a+(n-1)d]
S45 = 45/2 [2×10+(45-1)×2]
S45= 45/2 [20+44×2]
S45 = 45/2 (108)
S45 = 45×54
S45 = 2430