NCERT Solution
Class 11
Mathematics
Sequence and Series
find four numbers forming a geometric progression...

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.

Answer:

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

a₁ = a, a₂ = ar, a₃ = ar², a₄ = 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

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

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

4a = a + 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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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.

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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.

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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.