Find the G.P. whose sum upto infinite term is 3 and second term is -18.

Question

Ans .9,-18,36....... This is from the chapter geometric progression.

Answer 1

S=a+ar+ar^2+...=a(1-r^n)/(1-r). When n approaches infinity and -1<r<1, 3=a/(1-r), 3-3r=a. Also, ar=-18, so a=-18/r.

Therefore, 3-3r=-18/r, 3r-3r^2+18=0; 3r^2-3r-18=0=3(r+2)(r-3), r=-2, 3 and a=9, -6. However, r<-1 or r>1 and r^n itself approaches infinity, which suggests that the sum to infinity cannot be 3. S=(9/3)(1-(-2)^n)=3(1-(-2)^n). The series is 9-18+36-72+144-288 ... which doesn't appear to converge! Certainly the alternative series -6-18-54 doesn't!