Question

If \[a,b,c\] are in G.P., \[a - b,c - a,b - c\] are in H.P., then \[a + 4b + c\] is equal to
A.\[0\]
B.\[11\]
C.\[ - 1\]
D. None of these

Answer 1

Hint:
We review the terms arithmetic progression (AP), geometric progression (GP), and harmonic progression (HP) (HP). Using the definitions of Arithmetic progression, Geometric progression, and Harmonic progression, we find the standard relation of three consecutive terms a, b, c. To prove the statement, we compare them to the standard relationship of Arithmetic progression, Geometric progression, and Harmonic progression.
Formula used:
if \[a,b,c\], are GP
\[{b^2} = ac\]
\[{\rm{a}} - {\rm{b}},{\rm{c}} - {\rm{a}},{\rm{b}} - {\rm{c}}\] are H.P
\[\frac{1}{{a - b}},\frac{1}{{c - a}},\frac{1}{{b - c}}\] are in Arithmetic progression.
Complete step-by-step solution:
We know that a harmonic sequence, also known as harmonic progression, abbreviated as HP, is the sequence of reciprocals of terms in an arithmetic sequence, which means that if \[{x_1},{x_2},{x_3},......\] is an HP, and then \[\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},,......\] is in AP.
The terms \[a,b,c\] are in Geometric progression
Now, we have to wrote the above statement in equation form, we get
\[{b^2} = ac\]
Thus, we conclude that,\[{\rm{a}} - {\rm{b}},{\rm{c}} - {\rm{a}},{\rm{b}} - {\rm{c}}\] are in Harmonic Progression.
So, the reciprocals of the above mentioned expressions are in Arithmetic progression.
Now, it will be \[\frac{1}{{a - b}},\frac{1}{{c - a}},\frac{1}{{b - c}}\] are in Arithmetic progression.
Now, we have to solve for \[(c - a)\], we get
\[(c - a) = \frac{{2(a - b)(b - c)}}{{(a - c)}}\]
Now, we have to square either side of the equation to solve the fractional equation above, we obtain
\[ - {(c - a)^2} = 2\left( {ab - {b^2} - ac + bc} \right)\]
Solve the terms inside the parentheses, to make it simpler:
\[ - {(c - a)^2} = 2((a + c)b - 2ac)\]
Solve the left side of the equation using square formula:
\[ - {c^2} - {a^2} + 2ac = 2(a + c)b - 4ac\]
Now, we have to group the like terms for easy simplification:
\[ - {c^2} - {a^2} + 6ac = 2(a + c)b\]
Now, we have to restructure the above equation explicitly to have all the terms on one side, we have \[ - {c^2} - {a^2} - 2ac + 8ac - 2(a + c)b = 0\]
We have to simplify the above equation using square formula:
\[ - {(c + a)^2} + 8ac - 2(a + c)b = 0\]
Rearrange the above equation to write in standard form of quadratic equation, we get \[{(c + a)^2} + 2(a + c)b - 8ac = 0\]
Solve the above with quadratic formula:
\[(c + a) = \frac{{ - 2b \pm \sqrt {4{b^2} + 32ac} }}{2}\]
Solve the numerator, we get
\[(a + c) = \frac{{ - 2b \pm 6b}}{2} = - 4b\]
Now substitute the obtained value, we have
\[a + 4b + c = 0\]
Therefore, if \[a,b,c\] are in G.P.,\[a - b,c - a,b - c\]are in H.P., then \[a + 4b + c\] is equal to \[a + 4b + c = 0\]
Hence, the option A is correct.
Note:
Students may make mistakes when selecting the options for the question. We can prove the same thing for GP with three consecutive terms\[\frac{f}{r},f,fr\] where the common ratio is “r“ and HP with three consecutive terms\[\frac{1}{{f - d}},\frac{1}{f},\frac{1}{{f + d}}\].