Properties of GP | Video Lesson on Properties of GP

Geometric progression is a progression in which the ratio between each term and the preceding term is always a constant. The ratio is termed the common ratio and is denoted by the letter “r”. The terms in a GP are non-zero. The first term is called the initial term and is denoted by “a”. A GP is represented by a_n = ar^n-1. A finite geometric progression consists of finite terms, whereas an infinite geometric progression has infinite terms.

10 Important Properties of GP

1) The common ratio of GP is not altered, even if all the terms of the GP are multiplied or divided by the same non-0 constant.

2) The reciprocal of the terms of a GP also forms a geometric progression with a common ratio that is the same as that of the reciprocal of the original GP.

3) If each term of a geometric progression consisting of a common ratio r is raised to some power (say k), then the geometric progression after the process has a common ratio r^k.

4) The product of the terms in a GP is at the same distance from the beginning to the end and is the same. It is equal to the product of the first and last terms.

5) If the numbers in a GP are selected at intervals that are regular, then the resulting geometric progression is also a GP.

6) If a GP has non-negative terms, then the logarithm of each term forms an AP and vice versa.

7) Consider 3 non-0 numbers (say a, b, c) exist if and only if it satisfies the condition b² = ac.

8) If the first term of a GP is a and the last term is l, then the product of all terms in that geometric progression is of the form (al)^n/2.

9) Let there be “n” quantities in GP with the common ratio “r”, and S_n is the sum to n terms; if two are taken at the same time, then the sum of their product is

[(r) / (r + 1)] [S_n] [S_n-1].

10] Consider the GP a^x_{1}, a^x_{2}, …….. A^x_{n}. Then, x₁, x₂ …… are in AP.

Solved Problems on GP

Problem 1: If x, 2x + 2, 3x + 3 are in GP, then the fourth term is

A) 27

B) −27

C) 13.5

D) −13.5

Solution: D

Given that x, 2x + 2, 3x + 3 are in GP.

Therefore, (2x + 2)²= x (3x + 3)

⇒ x²+ 5x + 4 = 0

⇒ (x + 4) (x + 1) = 0

⇒ x = − 1, − 4

Now, the first term a = x

The second term ar = 2 (x + 1)

⇒ r = 2 (x + 1) / (x) then 4^thterm = ar³= x [2 (x + 1) / x]³

= (8 / x²) (x + 1)³

Putting x = − 4,

T₄= (8 / 16) (−3)³= − 27 / 2

= −13.5

Problem 2: The sum of the first five terms of the series

\(\begin{array}{l}3+4\frac{1}{2}+6\frac{3}{4}+……\end{array} \)

will be

Solution:

\(\begin{array}{l}3+4\frac{1}{2}+6\frac{3}{4}+……..\\ =3+\frac{9}{2}+\frac{27}{4}+…..\\ =3+\frac{{{3}^{2}}}{2}+\frac{{{3}^{3}}}{4}+\frac{{{3}^{4}}}{8}+\frac{{{3}^{5}}}{16}+…..(in GP) \\ a=3,\ r=\frac{3}{2} , \text { then sum of the five terms} \\ {{S}_{5}}=\frac{a({{r}^{n}}-1)}{r-1}=\frac{3\left[ {{\left( \frac{3}{2} \right)}^{5}}-1 \right]}{\frac{3}{2}-1}\\ =\frac{1\left[ \frac{{{3}^{5}}}{32}-1 \right]}{\frac{1}{2}}\\ =6\left[ \frac{243-32}{32} \right]\\ =\frac{211\times 3}{16}\\ =\frac{633}{16}\\=39\frac{9}{16}\end{array} \)

Problem 3: If every term of a GP with positive terms is the sum of its two previous terms, then find the common ratio of the series.

Solution:

If the first term and common ratio of GP are respectively a and r, then under the condition,

\(\begin{array}{l}{{T}_{n}}={{T}_{n-1}}+{{T}_{n-2}}\\ \Rightarrow a{{r}^{n-1}}=a{{r}^{n-2}}+a{{r}^{n-3}}\\ \Rightarrow a{{r}^{n-1}}=a{{r}^{n-1}}{{r}^{-1}}+a{{r}^{n-1}}{{r}^{-2}}\\ \Rightarrow 1=\frac{1}{r}+\frac{1}{{{r}^{2}}}\\ \Rightarrow {{r}^{2}}-r-1=0\\ \Rightarrow r=\frac{1\pm \sqrt{1+4}}{2}=\frac{1+\sqrt{5}}{2}\\ \text { Taking only (+) sign } (\because \ r>1)\end{array} \)

Properties of GP – Video Lesson

JEE Maths

1,900

Frequently Asked Questions on GP

Q1

What is a GP?

A geometric progression (GP) is a sequence in which the ratio between each term and the preceding term is always a constant.

Q2

What is a common ratio in GP?

A common ratio is the ratio between any two consecutive terms in a geometric progression. It is denoted by the letter ‘r’.

Q3

Give the equation for the n^th term of a GP.

The n^th term of a GP is given by a_n = ar^n-1.

Q4

Give an example of a GP.

3, 9, 27, 81 is an example of GP.
Here, the first term, a = 3, common ratio, r = 9/3 = 3.

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