UPSKILL MATH PLUS

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In this section, let us look at the general formula to find the \(n^{th}\) term of a geometric progression.

We know that the general form of a geometric progression is \(a\), \(ar\), \(ar^2\),\(...ar^{n - 1}\) with the common ratio \(r\).

\(\text{Term } 1 = t_1 = a \times r^0 = a \times 1 = a\)

\(\text{Term } 2 = t_2 = a \times r^1 = ar\)

\(\text{Term } 3 = t_3 = a \times r^2 = ar^2\)

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\(\text{Term } n = t_{n} = a \times r^{n - 1} = ar^{n - 1}\)

Therefore, the general form of G.P is \(ar^{n - 1}\).

**Ratio between any two consecutive terms in a G.P**:

\(\frac{t_2}{t_1} = \frac{ar}{a} = r\)

\(\frac{t_3}{t_2} = \frac{ar^2}{ar} = r\)

\(\frac{t_n}{t_{n - 1}} = \frac{ar^n}{ar^{n - 1}} = \frac{ar^n}{a \times r^n \times r^{-1}} = \frac{1}{r{-1}} = r\)

Thus, the ratio between any two consecutive terms of a G.P is \(r\).

**Condition for three numbers to be in G.P**:

Let the three numbers \(a\), \(b\), \(c\) in G.P.

\(\text{First term }= a\)

\(\text{Second term }= b = ar\)

\(\text{Third term }= c = ar^2\)

\(ac = a \times ar^2 = a^2r^2 = (ar)^2 = b^2\)

Thus, three non-zeroes numbers are in G.P only if \(b^2 = ac\).

Important!

**1.**When the product of the three consecutive terms of a G.P is given, then we can consider the terms as \(\frac{a}{r}\), \(a\), \(ar\).

**2.**When the product of the first four terms of a G.P are given, then we can consider the terms as \(\frac{a}{r^3}\), \(\frac{a}{r}\), \(ar\), \(ar^3\).

**3.**When every term of a G.P is divided or multiplied by the same non-zero constant, the newly-formed sequence is also in G.P.