krit.club logo

Exploring some more Progressions - Infinite Geometric Progression-advanced

Grade 9CBSE

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

🔑Concepts

•

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.

•

An infinite GP is a sequence a,ar,ar2,ar3,…a, ar, ar^2, ar^3, \dots that continues indefinitely.

•

The sum of an infinite GP exists only if the absolute value of the common ratio is less than 11, i.e., ∣r∣<1|r| < 1 or −1<r<1-1 < r < 1. In this case, the series is called a convergent series.

•

If ∣r∣≥1|r| \geq 1, the sum of the infinite GP does not approach a finite number; such a series is called divergent.

•

Infinite Geometric Progressions are often used to convert recurring decimals into rational fractions (p/qp/q form).

📐Formulae

S∞=a1−rS_{\infty} = \frac{a}{1 - r}

Condition for convergence: ∣r∣<1\text{Condition for convergence: } |r| < 1

an=arn−1a_n = a r^{n-1}

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

💡Examples

Problem 1:

Find the sum of the infinite geometric progression: 9,3,1,13,…9, 3, 1, \frac{1}{3}, \dots

Solution:

In the given series, the first term a=9a = 9 and the common ratio r=39=13r = \frac{3}{9} = \frac{1}{3}. Since ∣r∣=∣13∣<1|r| = |\frac{1}{3}| < 1, the sum to infinity exists. Using the formula S∞=a1−rS_{\infty} = \frac{a}{1 - r}: S∞=91−13S_{\infty} = \frac{9}{1 - \frac{1}{3}} S∞=923S_{\infty} = \frac{9}{\frac{2}{3}} S∞=9×32=272=13.5S_{\infty} = 9 \times \frac{3}{2} = \frac{27}{2} = 13.5

Explanation:

Identify the first term aa and common ratio rr. Verify that ∣r∣<1|r| < 1 before applying the sum formula for infinite terms.

Problem 2:

The sum of an infinite GP is 1515 and the sum of their squares is 4545. Find the first term and the common ratio.

Solution:

Let the GP be a,ar,ar2,…a, ar, ar^2, \dots. The sum is: S∞=a1−r=15…(1)S_{\infty} = \frac{a}{1-r} = 15 \quad \dots (1) Squaring the terms, we get the series a2,a2r2,a2r4,…a^2, a^2r^2, a^2r^4, \dots with first term a2a^2 and common ratio r2r^2. Its sum is: a21−r2=45…(2)\frac{a^2}{1-r^2} = 45 \quad \dots (2) From (1), a=15(1−r)a = 15(1-r). Substituting this into (2): (15(1−r))2(1−r)(1+r)=45\frac{(15(1-r))^2}{(1-r)(1+r)} = 45 225(1−r)2(1−r)(1+r)=45\frac{225(1-r)^2}{(1-r)(1+r)} = 45 225(1−r)1+r=45\frac{225(1-r)}{1+r} = 45 5(1−r)=1+r  ⟹  5−5r=1+r  ⟹  6r=4  ⟹  r=235(1-r) = 1+r \implies 5 - 5r = 1 + r \implies 6r = 4 \implies r = \frac{2}{3} Substituting r=23r = \frac{2}{3} in (1): a=15(1−23)=15(13)=5a = 15(1 - \frac{2}{3}) = 15(\frac{1}{3}) = 5

Explanation:

Form two equations based on the sum of the original series and the sum of the series formed by squares. Solve the system of equations for aa and rr.

Problem 3:

Represent the recurring decimal 0.424242…0.424242\dots (or 0.42‾0.\overline{42}) as a fraction in the form pq\frac{p}{q}.

Solution:

The decimal 0.424242…0.424242\dots can be written as a sum: S=0.42+0.0042+0.000042+…S = 0.42 + 0.0042 + 0.000042 + \dots This is an infinite GP where: First term a=0.42=42100a = 0.42 = \frac{42}{100} Common ratio r=0.00420.42=1100=0.01r = \frac{0.0042}{0.42} = \frac{1}{100} = 0.01 Since ∣r∣<1|r| < 1, use the formula S∞=a1−rS_{\infty} = \frac{a}{1 - r}: S=0.421−0.01S = \frac{0.42}{1 - 0.01} S=0.420.99S = \frac{0.42}{0.99} S=4299S = \frac{42}{99} Dividing both by 33, we get S=1433S = \frac{14}{33}

Explanation:

Express the decimal as an infinite sum of terms that form a GP. Calculate aa and rr, then apply the S∞S_{\infty} formula to get the fraction.