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.

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Solution

$a_{2}, a_{3}, a_{6}$ of A.P. are in G.P.

$a_{2} = a + d$ ......... (i)

$a_{3} = a + 2 d$ ......... (ii)

$a_{6} = a + 5 d$ ......... (iii)

$\frac{a_{3}}{a_{2}} = \frac{a_{6}}{a_{3}}$

$\frac{a + 2 d}{a + d} = \frac{a + 5 d}{a + 2 d}$

$a^{2} + 4 d^{2} + 4 a d = a^{2} + 5 a d + a d + 5 d^{2}$

$4 d^{2} - 5 d^{2} = 6 a d - 4 a d$

$- d^{2} = 2 a d$

$- d = 2 a$

$2 a = - d$

$a_{2} = - \frac{d}{2} + d = \frac{d}{2}$

$a_{3} = - \frac{d}{2} + 2 d = \frac{3 d}{2}$

$a_{62} = - \frac{d}{2} + 5 d = \frac{9 d}{2}$

$r = \frac{a_{3}}{a_{2}} = \frac{a_{6}}{a_{3}}$

$= \frac{\frac{3 d}{2}}{\frac{d}{2}} = \frac{\frac{9 d}{2}}{\frac{3 d}{2}}$

$r = 3$

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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.

a2,a3,a6 of A.P. are in G.P. a2=a+d ......... (i) a3=a+2d ......... (ii) a6=a+5d ......... (iii) a3a2=a6a3 a+2da+d=a+5da+2d a2+4d2+4ad=a2+5ad+ad+5d2 4d2−5d2=6ad−4ad −d2=2ad −d=2a 2a=−d a2=−d2+d=d2 a3=−d2+2d=3d2 a62=−d2+5d=9d2 r=a3a2=a6a3 =3d2d2=9d23d2 r=3