Question 21

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.

Answer

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

Substituting the value of *r* in (1), we obtain

4*a *= *a* + 9

⇒ 3*a* = 9

∴ *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, and –24.

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Saiganesh
2017-11-13 13:12:06

Good but it is very happy

Rajesh
2016-05-25 12:43:58

If we take 4 numbers in GP are a/r^3,a/r,ar and ar^3.than plz solve it...

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