Question

If \[x,2x + 2,3x + 3,\] are in G.P., then the fourth term is
A. \[27\]
B. \[ - 27\]
C. \[13.5\]
D. \[ - 13.5\]

Answer 1

Hint: Given that the series is in GP so we apply the G.P formula to create a quadratic equation which is \[{b^2} = ac\] where b is the second term of the series, a is the first term and c is the third term of the series. After simplifying it, we obtain a variable value. We next assume that the fourth term belongs to a variable. Finally, we apply the formula once more to produce the desired result.

Formula used:
We have been using the following formulas:
1. \[{b^2} = ac\] where b is the second term, a is the first term and c is the third term of the series.
2. \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]

Complete step-by-step solution:
We are given that \[x,2x + 2,3x + 3\] are in G.P.
Now we know that if a, b, and c are in G.P. then \[{b^2} = ac\] where b is the second term, a is the first term and c is the third term of the series.
Now we apply this formula to given series, we get
\[{\left( {2x + 2} \right)^2} = x\left( {3x + 3} \right)\]
We know that \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
\[{\left( {2x} \right)^2} + {2^2} + 2 \times 2x \times 2 = x \times 3x + x \times 3\]
By simplifying the above equation, we get
\[
  4{x^2} + 4 + 8x = 3{x^2} + 3x \\
  4{x^2} - 3{x^2} + 8x - 3x + 4 = 0 \\
  {x^2} + 5x + 4 = 0 \\
  {x^2} + 4x + x + 4 = 0 \\
 \]
Further simplify, we get
\[
  x\left( {x + 4} \right) + 1\left( {x + 4} \right) = 0 \\
  \left( {x + 4} \right)\left( {x + 1} \right) = 0 \\
  x = - 4, - 1 \\
 \]
Now when \[x = - 1\]
The series becomes
 \[
   - 1,2\left( { - 1} \right) + 2,3\left( { - 1} \right) + 3.. \\
   - 1,0,0.. \\
 \]
Therefore, the fourth term of the series is 0.
Hence, it is not possible.
When \[x = - 4\]
The series becomes
\[
   - 4,2\left( { - 4} \right) + 2,3\left( { - 4} \right) + 3 \\
   - 4, - 6, - 9 \\
 \]
Now let us assume that the fourth term is \[y\]
So, the series becomes \[x,2x + 2,3x + 3,y\]
Now we take second, third, and fourth term of this series that is \[2x + 2,3x + 3,y\]
Again, we apply the formula \[{b^2} = ac\] on the above series, we get
 \[{\left( {3x + 3} \right)^2} = \left( {2x + 2} \right)y\]
Now apply algebraic identity and simplify it, we get
\[
  {\left( {3x} \right)^2} + {3^2} + 2 \times 3x \times 3 = 2x \times y + 2 \times y \\
  9{x^2} + 9 + 18x = 2xy + 2y \\
  9\left( {{x^2} + 2x + 1} \right) = 2y\left( {x + 1} \right) \\
  9{\left( {x + 1} \right)^2} = 2y\left( {x + 1} \right) \\
 \]
Now on further simplification, we get
\[
  9\left( {x + 1} \right) = 2y \\
  y = \dfrac{{9\left( {x + 1} \right)}}{2} \\
 \]
Now substitute the value of \[x = - 4\] in the above equation, we get
\[
  y = \dfrac{{9\left( { - 4 + 1} \right)}}{2} \\
   = 9 \times \left( { - \dfrac{3}{2}} \right) \\
   = - \dfrac{{27}}{2} \\
 \]
Therefore, the fourth term of the series is \[ - \dfrac{{27}}{2}\]or \[ - 13.5\]
Hence, option (D) is correct

Note: Students must be careful while applying the algebraic identity and simplifying it. Also, be careful while substituting the value of x in a given series as it is very important to find the series without errors because with the help of this series, we find the fourth term of the series.