then the second term a₂ = ar = ar^(2-1)

the third term a₃ = a₂ × r = (ar) × r = ar² = ar^(3-1)

the fourth term a₄ = a₃ × r = ar² × r = ar ³ = ar ^(4-1)

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Looking at the pattern we can say that n^th term a_n = ar ^n-1

So n^th term of a GP with first term ‘a’ and common ratio ‘r’ is given by a_n = ar ^n-1 .

Let us consider some examples

Example-20.Find the 20^th and n ^th term of the GP.

5 / 2 , 5 / 4 , 5 / 8 ......

Solution:Here a = 5 / 2 and r = 5 / 4 / 5 / 2 = 1 / 2

Then        a_{20 = ar^(20-1) = 5/2(1/2)¹⁹ = 5 / 2²⁰}

and        a_{n = ar^(n-1) = 5/2(1/2)^n-1 = 5 / 2ⁿ}

Example-21.Which term of the GP : 2, 2 √2 , 4 ..... is 128 ?

Solution:Here a = 2 r = 2√2 / 2 = √2

Let 128 be the n^th term of the GP.

Then a_n = ar ^n-1 = 128

2.( √2)^n-1 = 128

(√ 2)^n-1 = 64

(2)^{n-1 / 2} = 2⁶