Question

Find the infinite geometric sum of 80 + 10 +1.25 + ... to 5 decimal places past the decimal point.

Answer 1

first term.

a= 80

second term. a2=10

so Common ratio r=a2/a =10/80=1/8=0.125

checking for the validity of common ratio.

a2×r=10×1/8=10×0.125=1.25 =a3= Third term.

Hence it is true.

Now 4th Term =1.25/8=0.15625

It can be observed that 4TH term Has digit that is 5 places past the decimal point.

Hence, 0.15625 is the terminating number as per the question.

So the required sum is.

S=a+a2+a3+a4

->S=80+10+1.25+0.15625

->S= 91.40625

Hope it helps...

Regards,

Leukonov/Olegion.