Question

If second, third and sixth terms of an A.P. are consecutive terms of a G.P., write the common ratio of the G.P. 

Answer 1

\[\text{ Here, second term }, a_2 = a + d\]
\[\text{ Third term }, a_3 = a + 2d\]
\[\text{ Sixth term }, a_6 = a + 5d \]
\[\text{ As, a_2 , a_3 and a_6 are in G . P } . \]
\[ \therefore \text{ First term of G . P } . = a_2 = A = a + d\]
\[\text{ Second term of G . P } . = Ar = a + 2d\]
\[\text{ Third term of G . P }. = A r^2 = a + 5d \]
\[ \therefore \left( a + 2d \right)^2 = \left( a + d \right) \times \left( a + 5d \right)\]
\[ \Rightarrow a^2 + 4ad + 4 d^2 = a^2 + 6ad + 5 d^2 \]
\[ \Rightarrow 2ad + d^2 = 0\]
\[ \Rightarrow d(2a + d) = 0\]
\[ \Rightarrow d = 0 or 2a + d = 0\]
\[\text{ But }, d = 0 \text{ is not possible } . \]
\[ \therefore d = - 2a\]
\[ \therefore r = \frac{a + 2d}{a + d}\]
\[ \Rightarrow r = \frac{a + 2( - 2a)}{a + ( - 2a)}\]
\[ \Rightarrow r = \frac{3}{1} = 3\]
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