Geometric progressions
Algebra
Level
3
In a geometric progression of real numbers, the sums of the first two terms is 7, and the sum of the first six term is 91. Find the sum of the first four terms.
The answer is 28.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Se "a" é o termo inicial e "q" a razao da progressão, temos que a+aq = 7
Logo, a(1+q) = 7
Por outro lado, a+aq+...+aq^5=91
Logo, a(q^6-1)/(q-1) = 91
Portanto, (q^6-1)/[(1+q)(q-1)] = 13 -> q^6 - 1 = 13(q^2-1) -> q = 1 ou q = -1 ou q^4+q^2+1 = 13 -> q^2 = 3 ou q^2 = -4
Testando, gera solução apenas q^2 = 3, de onde obtemos que a+aq+aq²+aq³ = a(q^4-1)/(q-1) = 7(q^4-1)/(q^2-1) = 7(q^2+1) = 7(3+1) = 28.