Find the first geometric threshold index

Does “first exceeds 56” include a term equal to 56?

Geometric rule

\(\displaystyle u_0=9,\quad u_{n+1}=2u_n\)

Find the first index at which the sequence exceeds 70

Move through indices 0 to 5. Find equality with 56, then find the first value greater than 56. Compare the index with the number of multiplications

Which index first exceeds 56?

A threshold asks for the first index

A value that works is not enough: it must be the earliest allowed index. Build a table by repeated multiplication or evaluate the explicit rule at candidate indices. Keep non-negative integer indices

1. 1Translate the exact comparison Exceeds means strictly greater; falls below means strictly less. Equality does not satisfy either \(\displaystyle u_n>T\quad\text{or}\quad u_n<T\)
2. 2Start at index zero The initial value has index 0. The index counts multiplications, not the number of listed values \(\displaystyle u_n=aR^n\)
3. 3Check both neighbouring terms For these positive monotone sequences, show that the previous term fails and the proposed term succeeds. Check index 0 separately if it already satisfies the condition

Geometric rule

\(\displaystyle u_n=7\times2^n\)

For the displayed geometric rule, find the first index exceeding 56.

For a sequence starting at 7 and doubling at each index, how do “first reaches 56 or more” and “first exceeds 56” differ?

A decreasing geometric sequence

With a positive start and a ratio strictly between 0 and 1, the terms decrease. For start 96 and ratio one half, compare the values around 20.

1. 1First below Index 2 fails the condition; index 3 succeeds. Earlier values are larger, so none was below 20.

Geometric rule

\(\displaystyle u_n=96(1/2)^n\)

Find the first index at which this sequence falls below 20.

For a sequence starting at 80 and halving at each index, which index first falls below 10?

Why the neighbouring check works here

Positive geometric growth or decay moves in one direction. If the preceding term fails the threshold and the next term succeeds, all earlier terms also fail. State this monotone setting rather than applying the shortcut to every sequence

1. 1State the scope of the argument Positive starts with ratios greater than 1 increase, and positive starts with ratios between 0 and 1 decrease. Their monotonicity justifies the neighbouring-term check; that shortcut alone does not establish the first crossing for an oscillating sequence

A student finds a term above the threshold at a positive index and stops. What is still needed to establish the first crossing in this increasing sequence?

Do not round before comparing

For start 5 and ratio one half, the exact value at index 2 is 1.25. Rounding it to 1 would wrongly suggest it is below 1.2. Keep exact or sufficiently precise values until the threshold decision is made

1. 1Keep exact values for comparison Compare unrounded values with the threshold. Round a requested final term value only after identifying the crossing

Geometric rule

\(\displaystyle u_n=5(1/2)^n\)

Find the first index below 1.2. Do not round the terms before comparing

A sequence starts at index 0 with value 18 and doubles. What is the first non-negative index at which it exceeds 15?

A constant geometric sequence starts at index 0 with value 9 and has ratio 1. When does it first exceed 10?

Choose a method and report the right quantity

Iteration and a table are sufficient for this bounded bank; an explicit rule can check a candidate efficiently. A numerical final answer does not show which method was used. If a term value is also requested, give it separately from the index and round only as directed

1. 1No guaranteed crossing for every rule Check the starting condition and the sequence behaviour. An oscillating sequence may cross earlier, then return; checking only two late neighbours does not prove that crossing was first