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.
