Question

If \[{\rm{x}},{\rm{2x}} + {\rm{2}},{\rm{3x}} + {\rm{3}} \ldots \ldots \ldots \ldots \] are in G.P, then the fourth term is -13.5.
A. True
B. False

Answer 1

Hint: First look at the geometric progression definition. Now find the relation between the first two terms and next two terms. By that you get an equation with x terms on both sides. Now do cross multiplication. By this you get quadratic equations on both sides of the equation. By simplifying this equation you can cancel out a few terms. Now you get the quadratic equation = 0. Now do factorization for this equation and find the value of x. Thus you get the first three terms. By these you can find out the fourth term which is the required result.

Complete step by step solution:
Geometric Progression: The sequence of numbers where each term is found by multiplying the previous one with a fixed non-zero number called common ratio. If the first term is a and common ratio r. The sequence is \[a,{\rm{ ar, a}}{{\rm{r}}^{\rm{2}}}........................\]
Given geometric sequence in terms of x, can be written as:
\[x,{\rm{ 2x + 2, 3x + 3, }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{.}}\]
From the definition, let us assume the common ratio of this sequence as r, we get relation between \[{{\rm{1}}^{{\rm{st}}}}\] and \[{{\rm{2}}^{{\rm{nd}}}}\] term:
\[x{\rm{ = }}\left( {2x + 2} \right)r{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}\](1)
By using definition, previous assumption, we get relation between next two terms as:
\[\left( {2x + 2} \right){\rm{ = }}\left( {3x + 3} \right)r{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{.}}\](2)
By dividing the equations (1), (2), we get it as follows:
\[\dfrac{x}{{2x + 2}}{\rm{ = }}\dfrac{{2x + 2}}{{3x + 3}}\]
By cross multiplying the terms, we get it as follows:
\[x\left( {3x + 3} \right){\rm{ = }}\left( {2x + 2} \right)\left( {2x + 2} \right)\]
By subtracting \[\left( {2x + 2} \right)\left( {2x + 2} \right)\]and simplifying, we get it as follow:
\[3{x^2} + 3x - 4{x^2} - 4 - 8x{\rm{ = 0}}\]
By simplifying and taking “-“as common, we get it as below:
\[{x^2} + 5x + 4{\rm{ = 0}}\]
By factorization, we can write 5x as 4x + x, we get it as:
\[{x^2} + 4x + x + 4{\rm{ = 0}}\]
Now taking x common from first two terms, (x+4) whole common, we get it as:
\[\left( {x + 4} \right)\left( {x + 1} \right){\rm{ = 0}}\]
So, roots are \[x = - 1, - 4\]
By substituting x = -1, we get sequence as -1, 0, 0, ……..
By looking at the above sequence we say common ratio = 0. But by definition we can say the common ratio is a non-zero number. So, x = -1 is a wrong substitution.
By substituting x = -1, we get sequence as \[ - {\rm{4}}, - {\rm{6}}, - {\rm{9 }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.{\rm{ }}.\]
So, common ratio (r) \[ = {\rm{ }}\dfrac{{{2^{nd}}{\rm{ term}}}}{{{1^{st}}{\rm{ term}}}}{\rm{ = }}\dfrac{{ - 6}}{{ - 4}}{\rm{ = }}\dfrac{3}{2}\]
So, $4^{th}$ term \[ = {\rm{ r }} \times {\rm{ }}{{\rm{3}}^{{\rm{rd}}}}{\rm{ term = }}\dfrac{3}{2} \times \left( { - 9} \right){\rm{ = - 13}}{\rm{.5}}\]
Therefore the given statement is true.

Note: Be careful while taking the x proving that -1 doesn’t satisfy is very important. You can see 0 can be ratio from the first two terms but you must also see about the next two terms. Generally students see the first two terms and think that it may be possible but no, it is not possible. After getting the common ratio (r) take care of “-“sign in the sequence. Do it carefully.