Find the numerical values of the terms of the geometric sequences 2k+5, k-5, k-9

Question

Geometric sequences of 2k+5, k-5, k-9

Answer 1

If a is the first term of the GP, ar the second and ar^2 the third, we can write a=2k+5, ar=k-5 and ar^2=k-9.

ar/a=r=(k-5)/(2k+5); but ar^2/ar=r=(k-9)/(k-5), so (k-5)/(2k+5)=(k-9)/(k-5).

Cross-multiplying: (k-5)^2=(k-9)(2k+5)=2k^2-13k-45. So k^2-10k+25=2k^2-13k-45 and k^2-3k-70=0, which factorises to (k-10)(k+7)=0. From this we have k=10 or -7. So a=2k+5=25 or -9. r=(k-5)/a=5/25=1/5 or 12/9=4/3. ar^2=25/25=10-9=1 (true); or ar^2=-9*16/9=-7-9=-16 (true). Both solutions for k appear valid, therefore a=10 and r=1/5; or a=-9 and r=4/3. The series is: 25, 5, 1, 0.2, etc., or -9, -12, -16, -64/3, etc.