Question

The first term of a geometric sequence is a and the common ratio is r. Given that, the third term of this sequence exceeds the second term by 20a. Find the value of r.​

Answer 1

`-4\ \mathrm{or\ }5.`

Explanation:

`\mathrm{Solution:}\\\mathrm{Given:}\\\mathrm{third\ term=second\ term+20a}\\\mathrm{or,\ }ar^2=ar+20a\\\mathrm{or,\ }r^2=r+20\ \ \ \ \ \ [\ a \\e 0\ ]\\\mathrm{or,\ }r^2-r-20=0\\\mathrm{or,\ }r^2+4r-5r-20=0\\\mathrm{or,\ }r(r+4)-5(r+4)=0\\\mathrm{or,\ }(r+4)(r-5)=0\\\mathrm{i.e.\ }r=-4\ \mathrm{or\ }5.`