SELECT * FROM question_mgmt as q WHERE id=1018 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=6 AND chapterId=80 and ex_no='3' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

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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