For a convergent GP, a=S_infinity(1-r). This converts a known total into an equation in the ratio. The finite sum satisfies S_n=S_infinity(1-rn), obtained by replacing a/(1-r) with S_infinity. Combine the supplied total with the other term or sum condition, solve all real candidates, check -1<r<1, then recover the first term. If S_infinity=20 and S_2=15, then r2=1-15/20=1/4. Both r=1/2,a=10 and r=-1/2,a=30 fit. Exam check: an even power does not choose the sign. A supplied “all terms positive” condition does; convergence must still be checked.

For a convergent GP, \(a=S_\infty(1-r)\). This converts a known total into an equation in the ratio.

The finite sum satisfies \(S_n=S_\infty(1-r^n)\), obtained by replacing \(\dfrac{a}{1-r}\) with \(S_\infty\).

Combine the supplied total with the other term or sum condition, solve all real candidates, check \(-1<r<1\), then recover the first term.

If \(S_\infty=20\) and \(S_2=15\), then \(r^2=1-\dfrac{15}{20}=\dfrac{1}{4}\). Both \(r=\dfrac{1}{2},a=10\) and \(r=-\dfrac{1}{2},a=30\) fit.

Exam check: an even power does not choose the sign. A supplied “all terms positive” condition does; convergence must still be checked.

Recovering an infinite progression

Recovering an infinite progression