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

Grade 11CBSE

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

🔑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,−4,8,−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.

📐Formulae

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

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

Sum of first nn terms (when r≠1r \neq 1): Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} or Sn=a(1−rn)1−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∞=a1−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.≥G.M.A.M. \geq G.M. (i.e., a+b2≥ab\frac{a+b}{2} \geq \sqrt{ab})

💡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.\n2. Find the 10th10^{th} term using an=arn−1a_n = ar^{n-1}:\na10=5⋅(2)10−1=5⋅29a_{10} = 5 \cdot (2)^{10-1} = 5 \cdot 2^9\na10=5⋅512=2560a_{10} = 5 \cdot 512 = 2560.\n3. Find the sum of the first 55 terms using Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} since r>1r > 1:\nS5=5(25−1)2−1=5(32−1)1S_5 = \frac{5(2^5 - 1)}{2 - 1} = \frac{5(32 - 1)}{1}\nS5=5⋅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.\n2. Here, a=3a = 3, n=4n = 4 (total terms), and a4=81a_4 = 81.\n3. Use the formula an=arn−1a_n = ar^{n-1} for the 4th4^{th} term:\n81=3⋅r4−1  ⟹  81=3⋅r381 = 3 \cdot r^{4-1} \implies 81 = 3 \cdot r^3\nr3=813=27r^3 = \frac{81}{3} = 27\nr=273=3r = \sqrt[3]{27} = 3.\n4. Find the missing terms:\nG1=a⋅r=3⋅3=9G_1 = a \cdot r = 3 \cdot 3 = 9\nG2=a⋅r2=3⋅32=27G_2 = a \cdot r^2 = 3 \cdot 3^2 = 27.\n5. 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.