A sequence is such that $u_{n+1} = u_n + 1.5$, for $n = 1, 2, 3, \dots$ Given that \(u_{50} = 99

Question

A sequence is such that $u_{n+1} = u_n + 1.5$, for $n = 1, 2, 3, \dots$ Given that $u_{50} = 99u_1$, show that $u_1 = 0.75$.[2] Find the least value of $n$ such that $u_2 + u_4 + u_6 + \dots + u_{2n}$ is more than $2025$.[2] It is given that $u_{25}$, $u_k$ and $u_5$ are the first three terms of a geometric progression. Find $k$.[2] Find the sum of the first $13$ terms of the geometric series, giving your answer to $3$ decimal places.[2]

Accepted Answer

(a) $u_1=0.75$ (b) $n=37$ (c)(i) $k=11$ (c)(ii) $64.311$

2025 NYJC P1 Q8

2025 NYJC P1 Q8

Solution:

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