There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6. If the first and the last numbers are equal then two other numbers are
A.
$$- 2, 4$$
B.
$$- 4, 2$$
C.
$$2, 6$$
D.
None
Answer :
$$- 4, 2$$
Solution :
Let the last three numbers in A.P. be $$a, a + 6, a + 12,$$ then the first term is also $$a + 12.$$
But $$a + 12, a, a + 6$$ are in G.P.
$$\eqalign{
& \therefore {a^2} = \left( {a + 12} \right)\left( {a + 6} \right) \cr
& \Rightarrow {a^2} = {a^2} + 18a + 72 \cr
& \therefore a = - 4. \cr} $$
∴ The numbers are $$8, - 4, 2, 8.$$
Releted MCQ Question on Algebra >> Sequences and Series
Releted Question 1
If $$x, y$$ and $$z$$ are $${p^{{\text{th}}}},{q^{{\text{th}}}}\,{\text{and }}{r^{{\text{th}}}}$$ terms respectively of an A.P. and also of a G.P., then $${x^{y - z}}{y^{z - x}}{z^{x - y}}$$ is equal to:
If $$a, b, c$$ are in G.P., then the equations $$a{x^2} + 2bx + c = 0$$ and $$d{x^2} + 2ex + f = 0$$ have a common root if $$\frac{d}{a},\frac{e}{b},\frac{f}{c}$$ are in-