Answer: Proved: \(t_{29} = 20d = 2 \times 10d = 2t_{19}\).
\(t_9 = a + 8d = 0\), so \(a = -8d\).
\[t_{29} = a + 28d = -8d + 28d = 20d\]\[t_{19} = a + 18d = -8d + 18d = 10d\]Hence \(t_{29} = 20d = 2(10d) = 2t_{19}\).
1 answer
Doubtly AI
· 24 Oct 2026
AI-verified · Oct 2026
Report
Was this answer helpful?
Thanks for the feedback.
Sign up free to save answers and ask your own doubts
Your answer
Log in or create a free account to answer this question.