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Sequences and Series - Geometric Progression (G.P.)

Grade 11CBSE

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

πŸ”‘Concepts

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A Geometric Progression (G.P.) is a sequence where each term after the first is obtained by multiplying the preceding term by a fixed, non-zero number called the common ratio (rr). Visually, if you plot the terms of a G.P. on a graph where the x-axis is the position nn and the y-axis is the value ana_n, the points will lie on an exponential curve rather than a straight line.

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The common ratio (rr) is the constant factor found by dividing any term by its immediate predecessor, i.e., r=a2a1=a3a2r = \frac{a_2}{a_1} = \frac{a_3}{a_2}. If r>1r > 1, the sequence shows exponential growth, moving rapidly away from the x-axis. If 0<r<10 < r < 1, the sequence shows exponential decay, gradually flattening and approaching the x-axis as nn increases.

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The general term or nthn^{th} term of a G.P. is denoted by ana_n. It allows us to find any specific term in the sequence without listing all previous terms. If the terms alternate in sign (e.g., 2,βˆ’4,8,βˆ’162, -4, 8, -16), the common ratio rr is negative, and the graph of the sequence would visually oscillate above and below the x-axis.

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The sum of the first nn terms (SnS_n) represents the total value of the series up to that point. The formula used depends on whether ∣r∣|r| is greater than or less than 11 to maintain positive denominators for ease of calculation, though both versions are mathematically equivalent.

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An infinite geometric series has a finite sum only if the absolute value of the common ratio is less than 11 (∣r∣<1|r| < 1). Visually, this means the terms become smaller and smaller, effectively 'vanishing' as nn approaches infinity, allowing the total sum to converge to a specific horizontal limit.

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The Geometric Mean (G.M.) between two positive numbers aa and bb is a number GG such that a,G,ba, G, b form a G.P. Geometrically, if aa and bb are the sides of a rectangle, the G.M. is the side of a square with the same area. The relationship is expressed as G=abG = \sqrt{ab}.

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To insert nn geometric means G1,G2,…,GnG_1, G_2, \dots, G_n between two numbers aa and bb, we create a G.P. with n+2n+2 terms where aa is the first term and bb is the (n+2)th(n+2)^{th} term. This is useful for interpolating values that follow a multiplicative trend between two known endpoints.

πŸ“Formulae

Common Ratio: r=ak+1akr = \frac{a_{k+1}}{a_k}

General Term (nthn^{th} term): an=arnβˆ’1a_n = a r^{n-1}

Sum of first nn terms (when rβ‰ 1r \neq 1): Sn=a(rnβˆ’1)rβˆ’1S_n = \frac{a(r^n - 1)}{r - 1} or Sn=a(1βˆ’rn)1βˆ’rS_n = \frac{a(1 - r^n)}{1 - r}

Sum of first nn terms (when r=1r = 1): Sn=naS_n = na

Sum of an infinite G.P. (when ∣r∣<1|r| < 1): S∞=a1βˆ’rS_{\infty} = \frac{a}{1 - r}

Geometric Mean (GG) of aa and bb: G=abG = \sqrt{ab}

Relationship between Arithmetic Mean (A.M.) and Geometric Mean (G.M.): A.M.β‰₯G.M.A.M. \geq G.M. (i.e., a+b2β‰₯ab\frac{a+b}{2} \geq \sqrt{ab})

πŸ’‘Examples

Problem 1:

Find the 10th10^{th} term and the sum of the first 55 terms of the G.P.: 5,10,20,40,…5, 10, 20, 40, \dots

Solution:

  1. Identify the first term and common ratio: First term a=5a = 5. Common ratio r=105=2r = \frac{10}{5} = 2.
  2. Find the 10th10^{th} term using an=arnβˆ’1a_n = ar^{n-1}: a10=5β‹…(2)10βˆ’1=5β‹…29a_{10} = 5 \cdot (2)^{10-1} = 5 \cdot 2^9 a10=5β‹…512=2560a_{10} = 5 \cdot 512 = 2560.
  3. Find the sum of the first 55 terms using Sn=a(rnβˆ’1)rβˆ’1S_n = \frac{a(r^n - 1)}{r - 1} since r>1r > 1: S5=5(25βˆ’1)2βˆ’1=5(32βˆ’1)1S_5 = \frac{5(2^5 - 1)}{2 - 1} = \frac{5(32 - 1)}{1} S5=5β‹…31=155S_5 = 5 \cdot 31 = 155.

Explanation:

We first identify the parameters aa and rr from the sequence. Since the ratio is constant (22), it is a G.P. We then apply the standard formulas for the general term and the finite sum.

Problem 2:

Insert two numbers between 33 and 8181 so that the resulting sequence is a G.P.

Solution:

  1. Let the two numbers be G1G_1 and G2G_2. The sequence is 3,G1,G2,813, G_1, G_2, 81.
  2. Here, a=3a = 3, n=4n = 4 (total terms), and a4=81a_4 = 81.
  3. Use the formula an=arnβˆ’1a_n = ar^{n-1} for the 4th4^{th} term: 81=3β‹…r4βˆ’1β€…β€ŠβŸΉβ€…β€Š81=3β‹…r381 = 3 \cdot r^{4-1} \implies 81 = 3 \cdot r^3 r3=813=27r^3 = \frac{81}{3} = 27 r=273=3r = \sqrt[3]{27} = 3.
  4. Find the missing terms: G1=aβ‹…r=3β‹…3=9G_1 = a \cdot r = 3 \cdot 3 = 9 G2=aβ‹…r2=3β‹…32=27G_2 = a \cdot r^2 = 3 \cdot 3^2 = 27.
  5. The sequence is 3,9,27,813, 9, 27, 81.

Explanation:

To insert terms, we treat the start and end values as the first and last terms of a G.P. We solve for the common ratio rr using the total number of terms, then multiply the first term progressively by rr to find the intermediate values.

Geometric Progression (G.P.) Class 11 Notes & Examples