# Find four numbers forming a G.P in which the third term is greater than the first term by 9, the second term is greater than the 4^{th} by 18

**Solution:**

Let a be the first term and r be the common ratio of the G.P. Hence,

a_{1} = a, a_{2} = ar, a_{3} = ar^{2}, a_{4} = ar^{3}

According to the given condition,

⇒ a_{3} = a_{1} + 9

⇒ ar^{2} = a + 9

⇒ ar^{2} - a = 9

⇒ a (r ^{2} - 1) = 9 ....(1)

⇒ a_{2} = a_{4} + 18

⇒ ar = ar^{3} + 18

⇒ ar^{3} - ar = - 18

⇒ ar (r ^{2} - 1) = -18 ....(2)

Dividing (2) by (1) , we obtain

⇒ ar (r ^{2} - 1)/a (r^{2} - 1) = - 18/9

⇒ r = - 2

Substituting r = - 2 in (1), we obtain

⇒ a [(- 2)2 - 1] = 9

⇒ a [4 - 1] = 9

⇒ 3a = 9

⇒ a = 9/3

⇒ a = 3

Thus, the first four numbers of the G.P. are 3, 3(- 2), 3(- 2)^{2,} and 3(- 2)^{3}

i.e., 3, - 6, 12, - 24

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 21

## Find four numbers forming a G.P in which the third term is greater than the first term by 9, the second term is greater than the 4^{th} by 18.

**Summary:**

We found the four numbers forming a G.P in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18 and the terms are 3, - 6, 12, - 24

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