Sarthaks Test
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Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

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Solution:

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

a1 = aa2 = ara3 = ar2a4 = ar3

By the given condition,

a3 = a1 + 9

⇒ ar2 = a + 9 … (1)

a2 = a4 + 18

⇒ ar ar3 + 18 … (2)

From (1) and (2), we obtain

a(r2 ­­– 1) = 9 … (3)

ar (1– r2) = 18 … (4)

Dividing (4) by (3), we obtain

fraction numerator a r open parentheses 1 minus r squared close parentheses over denominator a open parentheses r squared minus 1 close parentheses end fraction equals 18 over 9rightwards double arrow minus r space equals space 2rightwards double arrow r space equals space minus 2

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

4a + 9

⇒ 3a = 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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