Question 11

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Answer

Let the G.P. be T1, T2, T3, T4, … T2n.

Number of terms = 2n

According to the given condition,

T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1]

⇒ T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0

⇒ T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1]

Let the G.P. be aarar2ar3, …

therefore space fraction numerator a r space open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fraction space equals space fraction numerator 4 space cross times space a open parentheses r to the power of n space minus 1 close parentheses over denominator r space minus 1 end fraction
rightwards double arrow a r space equals space 4 a
rightwards double arrow r space equals space 4

Thus, the common ratio of the G.P. is 4.

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