Let's denote the terms of the geometric progression (G.P.) as follows:
a,ar,ar2,ar3,...,arn−1, where
a is the first term and
r is the common ratio. Given that the progression has an even number of terms, let's assume the total number of terms is
2n.
The sum of all the terms in the G.P. is given by the formula for the sum of a geometric series:
Now, consider the terms occupying the odd places. These terms are:
a,ar2,ar4,...,ar2n−2. The number of these terms is
n. Therefore, the sum of the terms in the odd places is:
Given that the sum of all the terms is 5 times the sum of the terms occupying odd places, we have the equation:
a‌=5(a‌) We can cancel out
a and
r2n−1 from both sides of the equation:
‌=‌ Multiplying both sides by
r2−1, we get:
r2−1=5(r−1)Expanding and simplifying the equation:
‌r2−1=5r−5‌r2−5r+4=0 To solve this quadratic equation, we can factorize it:
(r−1)(r−4)=0This gives us two possible values for
r :
r=1 or
r=4. However, since
r=1 would mean all terms are the same and do not form a progression, the valid solution is:
r=4Therefore, the common ratio of the G.P. is Option D: 4.