Six positive numbers are in GP, such that their product is 1000. If the fourth term is 1, then the last term is
1000
100
Five numbers are in AP with common difference 0. If the 1st, 3rd and 4th terms are in GP, then
the 5th term is always 0.
the 1st term is always 0.
the middle term is always 0
the middle term is always - 2.
Let f : R R be such that f is injective and f(x) f(y) = f(x + y) for all x, y if f(x), f(y) and f(z) are in GP, then x, y and z are in
AP always
GP always
AP depending on the values of x, y and z
GP depending on the values of x, y and z
A.
AP always
Let the function, f(x) = akx
Which define in f : R R and injective also.
Now, we have
f(x)f(y) = f(x + y)
On comparing, we get
2ky = k(x + z)