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Exploring some more Progressions - Sum of the first n terms of a Geometric Progression-advanced

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 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 denoted by aa and the common ratio is denoted by rr.

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The sum of the first nn terms (SnS_n) of a GP depends on the value of the common ratio rr.

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If r=1r = 1, the progression becomes a,a,a,…a, a, a, \dots and the sum of nn terms is simply n×an \times a.

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For an infinite geometric progression where ∣r∣<1|r| < 1, the sum to infinity (S∞S_{\infty}) exists because the terms get progressively smaller, approaching zero.

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The relationship between the nthn^{th} term (ana_n) and the sum of terms is given by an=Sn−Sn−1a_n = S_n - S_{n-1}.

📐Formulae

Sn=a(rn−1)r−1 when r>1S_n = \frac{a(r^n - 1)}{r - 1} \text{ when } r > 1

Sn=a(1−rn)1−r when r<1S_n = \frac{a(1 - r^n)}{1 - r} \text{ when } r < 1

Sn=na when r=1S_n = na \text{ when } r = 1

S∞=a1−r for ∣r∣<1S_{\infty} = \frac{a}{1 - r} \text{ for } |r| < 1

an=arn−1a_n = ar^{n-1}

💡Examples

Problem 1:

Find the sum of the first 66 terms of the geometric progression 5,10,20,40,…5, 10, 20, 40, \dots.

Solution:

Given the GP: 5,10,20,40,…5, 10, 20, 40, \dots First term a=5a = 5 Common ratio r=105=2r = \frac{10}{5} = 2 Number of terms n=6n = 6 Since r>1r > 1, we use the formula: Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} S6=5(26−1)2−1S_6 = \frac{5(2^6 - 1)}{2 - 1} S6=5(64−1)1S_6 = \frac{5(64 - 1)}{1} S6=5×63=315S_6 = 5 \times 63 = 315

Explanation:

We identify the first term and the common ratio. Since the common ratio r=2r=2 is greater than 11, we apply the sum formula for r>1r > 1 to find the total of the first 66 terms.

Problem 2:

In a GP, the first term is 77, the last term is 448448, and the sum is 889889. Find the common ratio rr.

Solution:

Given: a=7a = 7 an=448a_n = 448 Sn=889S_n = 889 We know an=arn−1a_n = ar^{n-1}, so 448=7×rn−1448 = 7 \times r^{n-1}, which means rn−1=64r^{n-1} = 64. Multiplying both sides by rr, we get rn=64rr^n = 64r. The sum formula is Sn=arn−ar−1S_n = \frac{ar^n - a}{r - 1}. Substituting the known values: 889=7(64r)−7r−1889 = \frac{7(64r) - 7}{r - 1} 889(r−1)=448r−7889(r - 1) = 448r - 7 889r−889=448r−7889r - 889 = 448r - 7 889r−448r=889−7889r - 448r = 889 - 7 441r=882441r = 882 r=882441=2r = \frac{882}{441} = 2

Explanation:

We use the relationship between the last term and the sum formula. By substituting arnar^n with an×ra_n \times r, we can solve for the common ratio rr directly without needing to find nn first.

Problem 3:

Find the sum of the infinite geometric series 1+13+19+…1 + \frac{1}{3} + \frac{1}{9} + \dots.

Solution:

Given the series: 1+13+19+…1 + \frac{1}{3} + \frac{1}{9} + \dots First term a=1a = 1 Common ratio r=1/31=13r = \frac{1/3}{1} = \frac{1}{3} Since ∣r∣<1|r| < 1, we use the formula for sum to infinity: S∞=a1−rS_{\infty} = \frac{a}{1 - r} S∞=11−13S_{\infty} = \frac{1}{1 - \frac{1}{3}} S∞=123=32=1.5S_{\infty} = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 1.5

Explanation:

This is an infinite GP with a common ratio less than 11. The sum approaches a finite limit, which is calculated using the formula for S∞S_{\infty}.