Question

If a, b, c are in AP, \[b - a\], \[c - b\] and \[a\] are in GP, then a: b: c is
(a) 1 : 2: 3
(b) 1: 3: 5
(c) 2: 3: 4
(d) 1: 2: 4

Answer 1

Hint: In this question, we need to find the ratio of a: b: c. The first step is to equate the common difference of the arithmetic progression i.e. \[d = {a_n} - {a_{n - 1}}\]. The common ratios of the GP are equated and simplified to get the required ratios.

Formula Used:
\[d = {a_n} - {a_{n - 1}}\]

Complete step by step Solution:
Given: a, b, c are in AP.
Here \[{a_1} = a\], \[{a_2} = b\] and \[{a_3} = c\]
We know that in arithmetic progression \[{a_2} - {a_1} = {a_3} - {a_2}\]
Equate the common difference of the sequence
\[b - a = c - b\] --- equation (1)
Solve for b
\[b = \dfrac{{a + c}}{2}\] ----- equation (2)
Now, \[b - a\],\[c - b\] and \[a\] are in GP
Geometric Progression (GP) is a type of sequence where each subsequent term is produced by multiplying each previous term by a fixed number called the common ratio. The common ratio of the geometric progression is given by the formula\[r = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\].
Since the common ratio \[r = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\]in GP, then the common ratios of the above sequence is\[\dfrac{{c - b}}{{b - a}} = \dfrac{a}{{c - b}}\]
Cross multiply the above equation
\[{(c - b)^2} = (b - a) \times a\]
Substitute \[b - a\] as \[c - b\] [from equation (1)]
\[{(c - b)^2} = (c - b) \times a\]
Solve for a
 \[(c - b) = a\]--- equation (3)
Substitute the value of equation (3) in equation (1)
\[b - a = a\]
\[b = 2a\] ---- equation (4)
Substitute the value\[b = 2a\] in equation (2)
\[2a = \dfrac{{a + c}}{2}\]
Solve for c to get the ratios of a:b:c
\[4a = a + c\]
\[4a - a = c\]
Therefore the value of c is
\[c = 3a\] -- equation (5)
 The ratios between a, b and c are a : 2a : 3a = 1:2:3

Hence, the correct option is a.

Note: The common difference in the arithmetic progression is given by the formula\[d = {a_n} - {a_{n - 1}}\]. Students should keep in mind the formula related to AP and GP to avoid mistakes.