What is the common ratio of the geometric sequence if the first term is 3 and its fifth term is 1875?

We will discuss here about the Geometric Progression along with examples.

A sequence of numbers is said to be Geometric Progression if the ratio of any term and its preceding term is always a constant quantity.

Definition of Geometric Progression:

A sequence of non-zero number is said to be in Geometric Progression (abbreviated as G.P.) if each term, after the first, is obtained by multiplying the preceding term by a constant quantity (positive or negative).

The constant ratio is said to be the common ratio of the Geometric Progression and is denoted by dividing any term by that which immediately precedes it.

In other words, the sequence {a\(_{1}\), a\(_{2}\), a\(_{3}\), a\(_{4}\), ..................., a\(_{n}\), ................. } is said to be in Geometric Progression, if \(\frac{a_{n + 1}}{a_{n}}\) = constant for all n ϵ N i.e., for all integral values of a, the ratio \(\frac{a_{n + 1}}{a_{n}}\) is constant.

Examples on Geometric Progression

1. The sequence 3, 15, 75, 375, 1875, .................... is a Geometric Progression, because \(\frac{15}{5}\) = \(\frac{75}{15}\) = \(\frac{375}{75}\) = \(\frac{1875}{375}\) = .................. = 5, which is constant.

Clearly, this sequence is a Geometric Progression with first term 3 and common ratio 5.

2. The sequence \(\frac{1}{3}\), -\(\frac{1}{2}\), \(\frac{3}{4}\), -\(\frac{9}{8}\), is a Geometric Progression with first term \(\frac{1}{3}\) and common ratio \(\frac{-\frac{1}{2}}{\frac{1}{3}}\) = -\(\frac{3}{2}\)

3. The sequence of numbers {4, 12, 36, 108, 324, ........... } forms a Geometric Progression whose common ratio is 3, because,

Second term (12) = 3 × First term (4),

Third term (36) = 3 × Second term (12),

Fourth term (108) = 3 × Third term (36),

Fifth term (324) = 3 × Fourth term (108) and so on.

In other words,

\(\frac{Second term (12)}{First term (4)}\) = \(\frac{Third term (36)}{Second term (12)}\) = \(\frac{Fourth term (108)}{Third term (36)}\) = \(\frac{Fifth term (324)}{Fourth term (108)}\) = ................. = 3 (a constant)

Solved example on Geometric Progression

Show that the sequence given by an = 3(2\(^{n}\)), for all n ϵ N, is a Geometric Progression. Also, find its common ratio.

Solution:

The given sequence is a\(_{n}\) = 3(2\(^{n}\))

Now putting n = n +1 in the given sequence we get,

a\(_{n + 1}\) = 3(2\(^{n + 1}\))

Now, \(\frac{a_{n + 1}}{a_{n}}\) = \(\frac{3(2^{n + 1})}{3(2^{n})}\) = 2

Therefore, we clearly see that for all integral values of n, the \(\frac{a_{n + 1}}{a_{n}}\) = 2 (constant). Thus, the given sequence is an Geometric Progression with common ratio 2.

Geometric Series:

If a\(_{1}\), a\(_{2}\), a\(_{3}\), a\(_{4}\), ..............., a\(_{n}\), .......... is a Geometric Progression, then the expression a\(_{1}\) + a\(_{2}\) + a\(_{3}\) + ......... + a\(_{n}\) + .................... is called a geometric series.

Notes:

(i) The geometric series is finite according as the corresponding Geometric Progression consists of finite number of terms.

(ii) The geometric series is infinite according as the corresponding Geometric Progression consists of infinite number of terms.

● Geometric Progression

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The fifth term of a G.P. is $$1875$$. If the first term is $$3$$, find the common ratio It is given that the first term of the G.P is $$a=3$$ and also the $$5$$th term is $$T_5=1875$$.

We know that the general term of an geometric progression with first term $$a$$ and common ratio $$r$$ is $$T_n=ar^{n-1}$$, therefore,

$$T_{ 5 }=ar^{ 5-1 }=ar^4$$

Now substitute $$a=3,n=5$$ and $$T_5=1875$$ in $$T_{ 5 }=ar^4$$ as follows:

$$T_{ 5 }=ar^{ 4 }\\ \Rightarrow 1875=3r^{ 4 }\\ \Rightarrow r^{ 4 }=\dfrac { 1875 }{ 3 } \\ \Rightarrow r^{ 4 }=625\\ \Rightarrow r^{ 4 }=5^{ 4 }\\ \Rightarrow r=5$$

Hence the common ratio is $$5$$.

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