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

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3$
$a_{2}, a_{3}, a_{6}$ are in $G . P .$

$\Rightarrow$ $a_{2} = a + d$

$\Rightarrow$ $a_{3} = a + 2 d$

$\Rightarrow$ $a_{6} = a + 5 d$

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

$\Rightarrow$ $a_{3}^{2} = a_{2} a_{6}$

$\Rightarrow$ ${(a + 2 d)}^{2} = (a + d) (a + 5 d)$

$\Rightarrow$ $a^{2} + {4 a}^{2} + 4 a d = a^{2} + 6 a d + {5 d}^{2}$

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

$\Rightarrow$ $d = - 2 a$

$∴$ $a_{2} = a + d = a - 2 a = - a a_{3} = a + 2 d = a - 4 a = - 3 a a_{6} = a + 5 d = a - 10 a = - 9 a$

$\Rightarrow$ $\frac{a_{3}}{a_{2}} = \frac{a_{6}}{a_{3}} = \frac{- 3 a}{- a} = \frac{- 9 a}{- 3 a} = 3$

Suggest Corrections

Similar questions

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.

Q. If

2^{n d}, 3^{r d}

and

6^{t h}

of an

A . P .

are three consecutive terms of a

G . P .

, then common ratio of the

G . P .

The sum of three numbers which are consecutive terms an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is

Q. If

p^{t h}, 2 p^{t h}, 4 p^{t h}

terms of an A.P. are in G.P. then common ratio of G.P. is

Join BYJU'S Learning Program

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

The correct option is B 3a2,a3,a6 are in G.P.⇒a2=a+d⇒a3=a+2d⇒a6=a+5d⇒a3a2=a6a3⇒a23=a2a6⇒(a+2d)2=(a+d)(a+5d)⇒a2+4a2+4ad=a2+6ad+5d2⇒−d2=2ad⇒d=−2a∴ a2=a+d=a−2a=−aa3=a+2d=a−4a=−3aa6=a+5d=a−10a=−9a⇒a3a2=a6a3=−3a−a=−9a−3a=3