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Class 11
Mathematics
Sequence and Series
the 5th 8th nbsp and 11th nbsp terms of a g p ar...

Question:
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

Answer:

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

According to the given condition,

a₅ = a r⁵–1 = a r⁴ = p … (1)

a₈ = a r⁸–1 = a r⁷ = q … (2)

a₁₁ = a r¹¹–1 = a r¹⁰ = s … (3)

Dividing equation (2) by (1), we obtain

$fraction numerator a r to the power of 7 over denominator a r to the power of 4 end fraction equals q over p r cubed equals q over p space space space space space space... left parenthesis 4 right parenthesis$

Dividing equation (3) by (2), we obtain

$fraction numerator a r to the power of 10 over denominator a r to the power of 7 end fraction equals s over q rightwards double arrow r cubed equals s over q space space space space space space space... left parenthesis 5 right parenthesis$

Equating the values of r3 obtained in (4) and (5), we obtain

$q over p equals s over q rightwards double arrow q squared equals p s$

Thus, the given result is proved.

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Question:
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

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Question: The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

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Question:
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.