Selecting Consecutive Terms in GP
Trending Questions
Q. The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, then the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is:
- 36
- 32
- 28
- 24
Q. In a G.P. if t3=2 and t6=−14, then t10=
- −1128
- 1128
- −164
- 164
Q. If f(x)=limn→∞(2x+4x3+…+2nx2n−1), where x∈(0, 1√2), then ∫f(x) dx is equal to
- log(1√1+2x2)+C
- log√1−2x2+C
- log(1√1−2x2)+C
- log(1−2x2)+C
Q. Let the digits of a three digit number are in G.P. If 400 is subracted from the number and the digits of the new number are in A.P., then the last digit of the original number is
- 8
- 9
- 1
- 2
Q. Let a, b, c be three distinct real numbers in geometric progression. If x is real and a+b+c=xb, then x can be
- −2
- 3
- −1
- 4
Q. The first term of a G.P. is 1. If the sum of the third and the fifth term is 90, then the common ratio is
- −√10
- √10
- −3
- 3
Q. Let an be the the nth term of a G.P. of positive numbers such that 100∑n=1a2n=α and 100∑n=1a2n−1=β, α≠β. Then the common ratio of G.P. is
- αβ
- βα
- √αβ
- √βα
Q. Three numbers are in G.P. If we double the middle term, then they will be in A.P. The common ratio of the G.P. is
- 1±√3
- 2±√3
- 1±√5
- 2±√5
Q. If x, G1, G2, y be the consecutive terms of a G.P., then the value of G1 G2 will be
- yx
- xy
- xy
- √xy
Q. Three positive numbers form an increasing G.P. If the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P. is
- 2−√3
- 2+√3
- √3−2
- √3
Q. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is:
- 2−√3
- 3+√2
- 2+√3
- √2+√3
Q. If three distinct real numbers a, b, c are in G.P and a+b+c=ax, then
- xϵ[34, ∞]
- xϵ R+
- x ϵ(−1, ∞)
- none of these