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Question -

Find four numbers forming a geometric progression in which third term isgreater than the first term by 9, and the second term is greater than the 4th by18.



Answer -

Consider a tobe the first term and r to be the common ratioof the G.P.

Then,

a1 = aa2 = ara3 = ar2a4 = ar3

From the question, wehave

a3 = a1 + 9

ar2 = a + 9 … (i)

a2 = a4 + 18

ar ar3 + 18 … (ii)

So, from (1) and (2),we get

a(r2 – 1) = 9 … (iii)

ar (1– r2) = 18 … (iv)

Now, dividing (4) by(3), we get

-r = 2

r = -2

On substituting thevalue of r in (i), we get

4a +9

3a = 9

 a =3

Therefore, the firstfour 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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