#### Question

Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and 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.

*a*_{1} = *a*, *a*_{2} = *ar*, *a*_{3} = *ar*^{2}, *a*_{4} = *ar*^{3}

By the given condition,

*a*_{3} = *a*_{1} + 9

⇒ *ar*^{2} = *a* + 9 … (1)

*a*_{2} = *a*_{4} + 18

⇒ *ar *= *ar*^{3} + 18 … (2)

From (1) and (2), we obtain

*a*(*r*^{2} – 1) = 9 … (3)

*ar *(1– *r*^{2}) = 18 … (4)

Dividing (4) by (3), we obtain

Is there an error in this question or solution?

Solution for question: Find Four Numbers Forming a Geometric Progression in Which Third Term is Greater than the First Term by 9, and the Second Term is Greater than the 4th by 18. concept: Geometric Progression (G. P.). For the courses CBSE (Commerce), CBSE (Science), CBSE (Arts)