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In a geometric sequence of real numbers, the sum of the first
two terms is 7 and the sum of the first six terms is 91. What is the sum of
the first four terms?
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We are given
+ = 7 (1)
+ + + + + = 91 (2)
In terms of "a" (the first term) and "r" (the common ratio) these equalities take the form
a + ar = 7, (3)
a + ar + + + + = 91 (4)
In (4), group the terms
(a + ar) + ( + ) + ( + ) = 91. (5)
Re-write (5) in an equivalent form
(a + ar) + + ) = 91. (6)
In (6), replace (a+ar) by the value of 7, based on (3). You will get
7 + + = 91. (7)
In (7), divide both sides by 7
1 + + = 13,
or
+ - 12 = 0. (8)
This biquadratic equation (8) is the quadratic equation relative . Solve it by factoring
= 0
Since we are given that the progression is in real numbers, can not be zero
(it is positive at any real r), we conclude that only possible value of is
= 3.
Then for the sum of the first four terms of this geometric progression we have
= a + ar + + = (a + ar) + ( + ) = (a+ar) + .
We substitute here a+ar = 7 and = 3, and we get
= 7 + 3*7 = 7 + 21 = 28.
ANSWER. The sum of the first 4 terms of this GP is 28.
Solved.