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Exploring some more Progressions - Introduction-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 of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero constant called the common ratio (rr).

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If the first term is aa and the common ratio is rr, the terms of the GP are a,ar,ar2,ar3,…,arn−1a, ar, ar^2, ar^3, \dots, ar^{n-1}.

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The nthn^{th} term (ana_n) is the general term used to find any specific position in the sequence.

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The sum of the first nn terms (SnS_n) of a GP depends on whether the common ratio rr is greater than, less than, or equal to 11.

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Special series include sequences where terms are powers of natural numbers, such as the sum of the first nn natural numbers, the sum of their squares, and the sum of their cubes.

📐Formulae

an=arn−1a_n = ar^{n-1}

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

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

∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2}

∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}

∑i=1ni3=[n(n+1)2]2\sum_{i=1}^{n} i^3 = \left[ \frac{n(n+1)}{2} \right]^2

💡Examples

Problem 1:

Find the 7th7^{th} term of the Geometric Progression 5,10,20,40,…5, 10, 20, 40, \dots

Solution:

Given: a=5a = 5, r=105=2r = \frac{10}{5} = 2. We need to find a7a_7. Using the formula an=arn−1a_n = ar^{n-1}: a7=5×(2)7−1a_7 = 5 \times (2)^{7-1} a7=5×26a_7 = 5 \times 2^6 a7=5×64=320a_7 = 5 \times 64 = 320.

Explanation:

Identify the first term and common ratio, then substitute into the general term formula for n=7n=7.

Problem 2:

Calculate the sum of the squares of the first 1010 natural numbers.

Solution:

We use the sum of squares formula: S=n(n+1)(2n+1)6S = \frac{n(n+1)(2n+1)}{6}. Here n=10n = 10. S=10(10+1)(2×10+1)6S = \frac{10(10+1)(2 \times 10 + 1)}{6} S=10×11×216S = \frac{10 \times 11 \times 21}{6} S=23106=385S = \frac{2310}{6} = 385.

Explanation:

Apply the specific summation formula for ∑n2\sum n^2 where nn is the total count of numbers.

Problem 3:

Find the sum of the first 66 terms of the GP 3,9,27,…3, 9, 27, \dots

Solution:

Given a=3a = 3 and r=93=3r = \frac{9}{3} = 3. Since r>1r > 1, use Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}. S6=3(36−1)3−1S_6 = \frac{3(3^6 - 1)}{3 - 1} S6=3(729−1)2S_6 = \frac{3(729 - 1)}{2} S6=3×7282S_6 = \frac{3 \times 728}{2} S6=3×364=1092S_6 = 3 \times 364 = 1092.

Explanation:

Identify aa and rr, determine which sum formula to use based on the value of rr, and calculate the result for n=6n=6.

Introduction-advanced Class 9 Notes & Examples | CBSE Maths