Question

obtain the sum 3+6+9+............+300​

Answer 1

Answer:

This type of questions first we need to find out which series it is AP or GP or HP series

So we can clearly see that it is AP series because all the term difference is same ie 3

As 6–3 = 3

9–6=3 & so on

So in AP series we have given

First term ie (a1= 3)

Difference ie (d= 3 )

Last term ie (an = 300)

So before finding sum we need to know number of terms ie n

an=a1+(n-1)d

300= 3+ (n-1)3

300–3=(n-1)3

297=(n-1)3

297/3=n-1

99=n-1

99+1=n

n=100

So now to find sum we have formula

Sn=n/2(2a1+(n-1)d)

Sn=100/2(2x3+(100–1)3)

Sn=50(6+99x3)

Sn=50(6+297)

Sn=50(303)

Sn=15150

Or we can solve in another way also

3+6+9+12+- - - - +300

3(1+2+3+4+ - - - - +100)

We know that sum of first n integers = n(n+1)/2

So 100(100+1)/2

50x101

5050

So we got sum of terms from 1 to 100 as 5050

Now multiply 3 with it

3x5050

So 15150

Answer 2

Step-by-step explanation:

here a=3

l=300

an=a+(n-1)d

300=3++n-1*3

297/3=n-1

99=n-1

n=100

sn= n/2*(a+(n-1)d)

=100/2*(2*3+(99)*3)

=50(303)

=15150