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Sequences and Progressions - Find nth term of geometric progressions and interpret GP growth patterns

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (rr).

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The first term of a GP is usually denoted by aa.

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The common ratio (rr) can be found by dividing any term by its preceding term: r=anan−1r = \frac{a_{n}}{a_{n-1}}.

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If ∣r∣>1|r| > 1, the sequence shows exponential growth (the values increase in magnitude).

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If 0<∣r∣<10 < |r| < 1, the sequence shows exponential decay (the values decrease towards zero).

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A sequence is a GP if the ratio a2a1=a3a2=⋯=anan−1\frac{a_2}{a_1} = \frac{a_3}{a_2} = \dots = \frac{a_n}{a_{n-1}} remains constant.

📐Formulae

an=a⋅rn−1a_n = a \cdot r^{n-1}

r=an+1anr = \frac{a_{n+1}}{a_n}

a,ar,ar2,ar3,…,arn−1a, ar, ar^2, ar^3, \dots, ar^{n-1}

💡Examples

Problem 1:

Find the 7th7^{th} term of the geometric progression: 5,10,20,40,…5, 10, 20, 40, \dots

Solution:

Identify the first term a=5a = 5. Calculate the common ratio r=105=2r = \frac{10}{5} = 2. We need to find the 7th7^{th} term (n=7n=7). Using the formula an=a⋅rn−1a_n = a \cdot r^{n-1}: a7=5⋅27−1a_7 = 5 \cdot 2^{7-1} a7=5⋅26a_7 = 5 \cdot 2^6 a7=5⋅64a_7 = 5 \cdot 64 a7=320a_7 = 320

Explanation:

To find a specific term, identify the starting value (aa) and the multiplier (rr), then apply the power n−1n-1 to the ratio.

Problem 2:

The population of a town triples every decade. If the initial population is 2,0002,000, what will the population be after 33 decades?

Solution:

The growth follows a GP where a=2000a = 2000 and r=3r = 3.

  • After 0 decades (Initial): a1=2000a_1 = 2000
  • After 1 decade: a2=2000⋅31a_2 = 2000 \cdot 3^1
  • After 3 decades: This corresponds to the 4th4^{th} term of the sequence (a4a_4). a4=2000⋅34−1a_4 = 2000 \cdot 3^{4-1} a4=2000⋅33a_4 = 2000 \cdot 3^3 a4=2000⋅27a_4 = 2000 \cdot 27 a4=54000a_4 = 54000

Explanation:

In growth patterns, 'after nn intervals' usually refers to the (n+1)th(n+1)^{th} term if the first term is the starting amount.

Problem 3:

Determine the common ratio and the nthn^{th} term expression for the GP: 81,27,9,3,…81, 27, 9, 3, \dots

Solution:

First term a=81a = 81. Common ratio r=2781=13r = \frac{27}{81} = \frac{1}{3}. The general nthn^{th} term is: an=81⋅(13)n−1a_n = 81 \cdot \left(\frac{1}{3}\right)^{n-1} Since 81=3481 = 3^4, we can simplify: an=34⋅13n−1a_n = 3^4 \cdot \frac{1}{3^{n-1}} an=34−(n−1)a_n = 3^{4-(n-1)} an=35−na_n = 3^{5-n}

Explanation:

The common ratio is less than 1, indicating this is a decaying GP. We used laws of exponents to simplify the final expression.