Question

find the sum of the geometric series 3+6+12+...+1536​

Answer 1

hope this helps you.. :)

Answer 2

The formula is........ tn=ar^(n-1)

&

sn= a.(r^n - 1)/(r-1)

Here, tn= 1536(because it is the last no)

a=3 (because,it started from 3)

r=2

(because3.2=6,6.2=12....768.2=1536)

sn= (3+6+12+.....1536)

♦At first we have to find the value of......

ar^(n-1) = 1536

or, 3.2^(n-1) =1536

or, 2^(n-1)= 1536/3

or, 2^(n-1)= 512

or, 2^(n-1)=2^9

or, n-1= 9

or, n= 9+1

or, n= 10

♦now put the value of......

sn=a.(r^n-1)/(r-1)

=3069(ans)