SEQUENCE AND SERIES: GEOMETRIC PROGRESSION (AP)
Content:
- Examples of geometric progression
- Calculation of; (i) First term (ii) Common ratio (iii) nth term, (iv) Geometric progression (v) sum of terms of geometric progression. (vi) Sum to infinity
- Practical problems involving real life situation.
GEOMETRIC PROGRESSION (G.P) OR EXPONENTIAL SEQUENCE
Given any sequence of terms T1, T2, T3, T4, … Tn-1, Tn. If the ratio between any term and the one preceding it is constant then the sequence is said to be in geometric progression (G.P). The ratio is called the common ratio denoted by r.
i.e. \(r = \frac{T_n}{T_{n – 1}}\)
where n = 1, 2, 3, 4, …
Example 1: Find out which of the sequence is a GP
(i) 1, 2, 4, 8, 16, …
\(\frac{T_n}{T_{n – 1}} ⇒ \frac{2}{1} = 2 \\ \frac{4}{2} = 2 \\ \frac{8}{4} = 2\)
The ratio is common, hence the sequence is a G.P ∴ r = 2
(ii) 16, 8, 4, 2, …
\(\frac{T_n}{T_{n – 1}} ⇒ \frac{8}{16} = \frac{1}{2} \\ \frac{4}{8} = \frac{1}{2} \\ \frac{2}{4} = \frac{1}{2}\)
The ratio is common, hence the sequence is a G.P ∴ \(r = \frac{1}{2}\)
Example 2: Find out which of the sequence is a GP
(i) 3, 7, 9, 12, …
\(\frac{T_n}{T_{n – 1}} ⇒ \frac{7}{3} = 2\frac{1}{3} \\ \frac{9}{7} = 1\frac{2}{7} \\ \frac{12}{9} = 1\frac{3}{9} = 1\frac{1}{3} \)
The ratio is NOT common, hence not a G.P.
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