Question

Suppose four distinct positive integers $a_1,a_2,a_3,a_4$ are in GP. Let $b_1=a_1$, $b_2=b_1+a_2$ , $b_3=b_2+a_3$ and $b_4=b_3+a_4$

Statement I : The numbers $b_1,b_2,b_3,b_4$ are neither in AP nor in GP.

Statement II : The numbers$b_1,b_2,b_3,b_4$ are in HP.

• A. Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I.
• B. Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I.
• C. Statement I is true; Statement II is false.
• D. Statement I is false; Statement II is true.

Answer 1

The correct option is C

Statement I is true; Statement II is false.

Explanation for the correct options:

Explanation for Statement-I : Finding if Statement I is true

Given $a_1,a_2,a_3,a_4$ are in GP.

Now

$$\begin{gathered} b_2-b_1=b_1+a_2-a_1 \\ =a_1+a_2-a_1 \\ =a_2 \end{gathered}$$

and

$$\begin{gathered} b_3-b_2=b_2+a_3-b_1-a_2 \\ =a_3 \end{gathered}$$

Thus, $b_2-b_1\ne b_3-b_2$. Hence, $b_1,b_2,b_3,b_4$ are not in AP.

Again,

$$\begin{gathered} b_1{b}_3=a_1\left(b_2+a_3\right) \\ =a_1\left(b_1+a_2+a_3\right) \\ =a_1\left(a_1+a_2+a_3\right) \\ ={a_1}^2+a_1{a}_2+a_1{a}_3 \end{gathered}$$

and

$$\begin{gathered} {b_2}^2={\left(b_1+a_2\right)}^2\left[\because b_2=b_1+a_2\right] \\ ={\left(a_1+a_2\right)}^2\left[\because b_1=a_1\right] \\ ={a_1}^2+2a_1{a}_2+{a_2}^2 \\ ={a_1}^2+2a_1{a}_2+a_1{a}_3\left[\because {a_2}^2=a_1{a}_3\right] \end{gathered}$$

Thus ${b_2}^2\ne b_1{b}_3$. Hence, $b_1,b_2,b_3,b_4$ are not in GP.

Therefore $b_1,b_2,b_3,b_4$ are neither in AP nor in GP.

Hence, statement I is true.

Explanation for Statement-II : Finding if Statement II is true

Formula to be used : We know that $b_1,b_2,b_3,b_4$ are in HP if $\frac{1}{b_1},\frac{1}{b_2},\frac{1}{b_3},\frac{1}{b_4}$ are in AP.

Now,

$$\begin{gathered} \frac{1}{b_2}-\frac{1}{b_1}=\frac{1}{b_1+a_2}-\frac{1}{a_1} \\ =\frac{1}{a_1+a_2}-\frac{1}{a_1} \\ =\frac{a_1-a_1-a_2}{a_1\left(a_1+a_2\right)} \\ =-\frac{a_2}{a_1\left(a_1+a_2\right)} \end{gathered}$$

Again,

$$\begin{gathered} \frac{1}{b_3}-\frac{1}{b_2}=\frac{1}{b_2+a_3}-\frac{1}{b_1+a_2} \\ =\frac{1}{a_1+a_2+a_3}-\frac{1}{a_1+a_2} \\ =\frac{a_1+a_2-a_1-a_2-a_3}{\left(a_1+a_2\right)\left(a_1+a_2+a_3\right)} \\ =-\frac{a_3}{\left(a_1+a_2\right)\left(a_1+a_2+a_3\right)} \end{gathered}$$

Thus, $\frac{1}{b_2}-\frac{1}{b_1}\ne \frac{1}{b_3}-\frac{1}{b_2}$. So, $b_1,b_2,b_3,b_4$ are not in HP.

Thus statement II is false.

Hence, Statement I is true and statement II is false.

Therefore, option (C) is the correct answer.