A real geometric sequence satisfies a₂+a₅=9 and a₇+a₁₀=288. Find a₂₀.

01 / Standard#Your turn

Official paper · jm01-2024 · I.7 · PDF 2

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

The second sum is r⁵ times the first.

Hint 2

Recover a₂ before moving to a₂₀.

Worked solution

Divide the two nonzero sums.
$r^5=288/9=32\implies r=2$
Find a₂ and advance eighteen places.
$a_2(1+2^3)=9\implies a_2=1,\quad a_{20}=2^{18}=262144$

D: 262144.

Checks and common pitfalls: From the second to the twentieth term there are eighteen ratios.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Divide the two nonzero sums.
$r^5=288/9=32\implies r=2$
Find a₂ and advance eighteen places.
$a_2(1+2^3)=9\implies a_2=1,\quad a_{20}=2^{18}=262144$

Think first. Reveal a hint when the class is ready.

2024 JM01