Question

Let a,b and c be in G.P with common ratio, where $a\ne 0$ and $0 < r\le \dfrac{1}{2}$. If 3a, 7b and 15c are in A.P then the fourth term of the A.P is
[a] $\dfrac{7a}{3}$
[b] a
[c] $\dfrac{2a}{3}$
[d] 5a

Answer 1

Hint: Use the fact that if a,b and c are in G.P then ${{b}^{2}}=ac$ and if a,b and c are in A.P then $a+c=2b$. Hence form two equations in a,b and c. Hence express b and c in terms of a and hence find the common difference of the A.P. Use the fact that if a is the first term of an A.P and d is the common difference term then the ${{n}^{th}}$ term of the A.P is given by ${{a}_{n}}=a+\left( n-1 \right)d$ and hence find the ${{4}^{th}}$ term of the A.P.

Complete step-by-step solution:
We know that if a,b and c are in G.P then ${{b}^{2}}=ac$
Hence, we have
${{b}^{2}}=ac\ \ \ \ \ \ \ \left( i \right)$
We know that if a,b and c are in A.P then $a+c=2b$.
Since 3a,7b and 15c are in A.P.
Hence, we have
$\begin{align}
  & 3a+15c=14b \\
 & \Rightarrow 14b=3a+15c\ \ \ \ \ \ \left( ii \right) \\
\end{align}$
Squaring equation (ii), we get
$196{{b}^{2}}={{\left( 3a+15c \right)}^{2}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using the above identity, we get
$196{{b}^{2}}=9{{a}^{2}}+90ac+225{{c}^{2}}$
Substituting the value of ${{b}^{2}}$ from equation (i), we get
$196ac=9{{a}^{2}}+90ac+225{{c}^{2}}$
Subtracting 196ac from both sides, we get
$225{{c}^{2}}-106ac+9{{a}^{2}}=0$
This is a quadratic equation in c.
We know that the solution of the quadratic equation $a{{x}^{2}}+bx+c=0$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here $a=225,b=-106a$ and $c=9{{a}^{2}}$
Hence, we have
$\begin{align}
  & c=\dfrac{106a\pm \sqrt{{{106}^{2}}{{a}^{2}}-4\left( 225 \right)\left( 9{{a}^{2}} \right)}}{2\times 225} \\
 & =\dfrac{106a\pm 56a}{450} \\
 & =\dfrac{9}{25}a,\dfrac{a}{9} \\
\end{align}$
When $c=\dfrac{9}{25}a$, we have
$\begin{align}
  & 14b=3a+\dfrac{15\left( 9 \right)}{25}a=\dfrac{42}{5}a \\
 & \Rightarrow b=\dfrac{3}{5}a \\
\end{align}$
Since $r\le \dfrac{1}{2}$, we have $\dfrac{b}{a}\le \dfrac{1}{2}$
Hence, we have $b=\dfrac{3a}{5}$ is rejected and hence $c=\dfrac{9}{25}a$ is rejected.
Hence, we have
$c=\dfrac{a}{9}$
Substituting the value of c in equation (ii), we get
$\begin{align}
  & 14b=3a+\dfrac{15a}{9}=\dfrac{14}{3}a \\
 & \Rightarrow b=\dfrac{a}{3} \\
\end{align}$
Hence, the A.P is
$3a,\dfrac{7a}{3},\dfrac{5a}{3}$
Hence the common difference if the A. P is $\dfrac{7a}{3}-3a=\dfrac{-2a}{3}$
Hence the fourth term of the A.P is $\dfrac{5a}{3}-\dfrac{2a}{3}=a$
Hence option [b] is correct

Note: In this question a student can make a mistake by not taking into consideration the bounds on the common ratio and hence will not reject the value of $c=\dfrac{9a}{25}$ which will lead to incorrect results. A common way to avoid this type of mistake is to check after the solution whether all the data of the question is used or not. If not then one should recheck his/her solution for possible mistakes.