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Algebra - Geometric Progression (GP)

Grade 10ICSE

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

🔑Concepts

A Geometric Progression (GP) is a sequence of non-zero 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). Visually, this creates a pattern of constant scaling rather than constant addition.

The Common Ratio (rr) is the constant multiplier. It is calculated by dividing any term by its preceding term, such as r=T2T1=T3T2r = \frac{T_2}{T_1} = \frac{T_3}{T_2}. If r>1r > 1, the sequence grows larger (diverges), while if 0<r<10 < r < 1, the terms get smaller (converges) towards zero.

The General Form of a GP is represented as a,ar,ar2,ar3,,arn1a, ar, ar^2, ar^3, \dots, ar^{n-1}, where aa is the first term and nn is the number of terms. On a graph, a GP does not form a straight line like an Arithmetic Progression; instead, it forms an exponential curve that becomes steeper as nn increases if r>1r > 1.

The General Term or nthn^{th} term (TnT_n) allows you to find any specific position in the sequence. It is defined as Tn=arn1T_n = a r^{n-1}. This formula shows that the exponent of the ratio is always one less than the position of the term.

When solving problems involving three numbers in a GP, it is often visually and algebraically simpler to assume the terms are ar,a,ar\frac{a}{r}, a, ar. This is because their product is simply a3a^3, as the rr terms cancel out, simplifying the calculation of the middle term.

The Sum of nn terms (SnS_n) represents the total value of all terms in the sequence up to nn. The formula changes slightly depending on whether rr is greater than or less than 1 to keep the denominator positive, but both forms are mathematically equivalent.

The Geometric Mean (GM) between two numbers aa and bb is a number GG such that a,G,ba, G, b are in GP. Algebraically, G2=abG^2 = ab or G=abG = \sqrt{ab}. Visually, the GM represents the side of a square that has the same area as a rectangle with sides aa and bb.

📐Formulae

Common Ratio: r=TnTn1r = \frac{T_n}{T_{n-1}}

General Term (nthn^{th} term): Tn=arn1T_n = a r^{n-1}

Sum of nn terms (when r>1r > 1): Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r - 1}

Sum of nn terms (when r<1r < 1): Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r}

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

Selection of 3 terms in GP: ar,a,ar\frac{a}{r}, a, ar

Selection of 4 terms in GP: ar3,ar,ar,ar3\frac{a}{r^3}, \frac{a}{r}, ar, ar^3 (common ratio r2r^2)

💡Examples

Problem 1:

Find the 9th9^{th} term of the Geometric Progression: 5,10,20,40,5, 10, 20, 40, \dots

Solution:

  1. Identify the first term (aa) and common ratio (rr): a=5a = 5 r=105=2r = \frac{10}{5} = 2
  2. Use the formula for the nthn^{th} term: Tn=arn1T_n = a r^{n-1}
  3. Substitute n=9n = 9, a=5a = 5, and r=2r = 2: T9=5×(2)91T_9 = 5 \times (2)^{9-1} T9=5×28T_9 = 5 \times 2^8
  4. Calculate the power and final product: 28=2562^8 = 256 T9=5×256=1280T_9 = 5 \times 256 = 1280

Explanation:

To find a specific term in a GP, we first determine the starting value and the multiplier. Since each term doubles, r=2r=2. Applying the general term formula gives us the value at the 9th position.

Problem 2:

Find the sum of the first 6 terms of the GP: 1,3,9,27,1, 3, 9, 27, \dots

Solution:

  1. Identify the given values: First term a=1a = 1 Common ratio r=31=3r = \frac{3}{1} = 3 Number of terms n=6n = 6
  2. Since r>1r > 1, use the formula: Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r - 1}
  3. Substitute the values: S6=1(361)31S_6 = \frac{1(3^6 - 1)}{3 - 1}
  4. Calculate 363^6: 36=7293^6 = 729
  5. Complete the calculation: S6=72912S_6 = \frac{729 - 1}{2} S6=7282=364S_6 = \frac{728}{2} = 364

Explanation:

We use the summation formula for r>1r > 1. By substituting the first term and the common ratio, we aggregate the values of the first six terms efficiently without adding them manually.