Question

Suppose 4 distinct positive integers ${a_1},{a_2},{a_3},{a_4}$ are in G.P. . Let ${b_1} = {a_1}{\text{, }}{b_2} = {b_1} + {a_2}{\text{, }}{{\text{b}}_3} = {b_2} + {a_3},{\text{ }}{{\text{b}}_4} = {b_3} + {a_4}$ , then;
Statement $\left( {\text{I}} \right)$ : The numbers ${b_1},{b_2},{b_3},{b_4}$ are neither in A.P. nor in G.P.
Statement $\left( {{\text{II}}} \right)$ : The numbers ${b_1},{b_2},{b_3},{b_4}$ are in H.P.
$\left( 1 \right)$ Statement I is true and statement II is also true ; statement II is the correct explanation of statement I .
$\left( 2 \right)$ Statement I is true and statement II is also true ; statement II is not the correct explanation of statement I .
$\left( 3 \right)$ Statement I is true and statement II is false
$\left( 4 \right)$ Statement I is false and statement II is true

Answer 1

In order to solve this question, we should be familiar with the basic properties of Geometric progression (G.P.) and Arithmetic progression (A.P.) . $\left( 1 \right)$ Geometric progression is a type of sequence in which each next term can be found out by multiplying the previous term with a constant number. It is generally represented like: \left\\{ {a,{\text{ }}ar,{\text{ }}a{r^2},{\text{ }}a{r^3},{\text{ }}a{r^4}..........} \right\\} where $a$ is the first term and $r$ is the common ratio which can be calculated as: $r = \dfrac{{{\text{Second term}}}}{{{\text{First term}}}}$ . $\left( 2 \right)$ Arithmetic progression is a type of sequence in which the difference of any two successive or consecutive numbers is always constant, that constant value is termed as a common difference and is denoted by $d$ . An arithmetic sequence in terms of common difference is represented as: \left\\{ {a,{\text{ }}a + d,{\text{ }}a + 2d,{\text{ }}a + 3d,...........a + \left( {n - 1} \right)d} \right\\} .

Complete step-by-step solution:
Since ${a_1},{a_2},{a_3},{a_4}$ are in G.P. ( given ) , then by the basic properties of G.P. ;
$\Rightarrow {a_1}{\text{ , }}{a_2} = {a_1}r{\text{ , }}{a_3} = {a_1}{r^2}{\text{ , }}{a_4} = {a_1}{r^3}$
Now, let us calculate the values of ${b_1},{b_2},{b_3},{b_4}$ in terms of ${a_1},{a_2},{a_3},{a_4}$ ;
$\because {b_1} = {a_1}{\text{ }}......\left( 1 \right)$
$\Rightarrow {b_2} = {b_1} + {a_2} = {a_1} + {a_2}$ (Putting the value of ${b_1}$ from equation $\left( 1 \right)$ )
Put the value of ${a_2} = {a_1}r$ , in the above equation we get;
$\Rightarrow {b_2} = {a_1} + {a_1}r = {a_1}\left( {1 + r} \right)$
$\therefore {b_2} = {a_1}\left( {1 + r} \right){\text{ }}......\left( 2 \right)$
$\Rightarrow {b_3} = {b_2} + {a_3} = {a_1} + {a_2} + {a_3}$ (Putting the value of ${b_2}$ from equation $\left( 2 \right)$ )
Put the value of ${a_3} = {a_1}{r^2}$ , in the above equation we get;
$\Rightarrow {b_3} = {a_1}\left( {1 + r} \right) + {a_1}{r^2}$
$\Rightarrow {b_3} = {a_1}\left( {1 + r + {r^2}} \right){\text{ }}......\left( 3 \right)$
$\Rightarrow {b_4} = {b_3} + {a_4} = {a_1} + {a_2} + {a_3} + {a_4}$ (Putting the value of ${b_3}$ from equation $\left( 3 \right)$ )
Put the value of ${a_4} = {a_1}{r^3}$ , in the above equation we get;
$\Rightarrow {b_4} = {a_1}\left( {1 + r + {r^2}} \right) + {a_1}{r^3}$
$\Rightarrow {b_4} = {a_1}\left( {1 + r + {r^2} + {r^3}} \right){\text{ }}......\left( 4 \right)$
Let us check for both the given statements now;
Statement $\left( {\text{I}} \right)$ : The numbers ${b_1},{b_2},{b_3},{b_4}$ are neither in A.P. nor in G.P.
$\left( 1 \right)$ If ${b_1},{b_2},{b_3},{b_4}$ are in A.P. then the below condition must be true;
$\Rightarrow {b_2} - {b_1} = {b_3} - {b_2}$ (must be true if in A.P.)
Substituting the values of ${b_1},{b_2},{b_3}$ from equation $\left( 1 \right),\left( 2 \right){\text{ and}}\left( 3 \right)$ ;
$\Rightarrow {a_1}\left( {1 + r} \right) - {a_1} = {a_1} + {a_1}r + {a_1}{r^2} - {a_1} - {a_1}r$
On further simplification;
$\Rightarrow {a_1}r \ne {a_1}{r^2}$
Therefore, ${b_1},{b_2},{b_3},{b_4}$ are not in A.P.
$\left( 2 \right)$ If ${b_1},{b_2},{b_3},{b_4}$ are in G.P. then the below condition must be true;
$\Rightarrow \dfrac{{{b_2}}}{{{b_1}}} = \dfrac{{{b_3}}}{{{b_2}}}$ (must be true if in G.P.)
$\Rightarrow \dfrac{{a\left( {1 + r} \right)}}{a} = \dfrac{{a\left( {1 + r + {r^2}} \right)}}{{a\left( {1 + r} \right)}}$
$\Rightarrow 1 + r \ne \dfrac{{1 + r + {r^2}}}{{1 + r}}$
Therefore, ${b_1},{b_2},{b_3},{b_4}$ are not in G.P.
Hence statement I is true that ${b_1},{b_2},{b_3},{b_4}$ are neither in A.P. nor in G.P.
Statement $\left( {{\text{II}}} \right)$ : The numbers ${b_1},{b_2},{b_3},{b_4}$ are in H.P.
If ${b_1},{b_2},{b_3},{b_4}$ are in H.P. then the below condition must be true;
$\Rightarrow \dfrac{1}{{{b_2}}} - \dfrac{1}{{{b_1}}} = \dfrac{1}{{{b_3}}} - \dfrac{1}{{{b_2}}}$ (must be true if in H.P.)
$\Rightarrow \dfrac{1}{{a\left( {1 + r} \right)}} - \dfrac{1}{a} = \dfrac{1}{{a\left( {1 + r + {r^2}} \right)}} - \dfrac{1}{{a\left( {1 + r} \right)}}$
On further simplification;
$\Rightarrow \dfrac{1}{a}\left( {\dfrac{1}{{1 + r}} - 1} \right) = \dfrac{1}{a}\left( {\dfrac{1}{{\left( {1 + r + {r^2}} \right)}} - \dfrac{1}{{\left( {1 + r} \right)}}} \right)$
Further simplifying the above equation;
$\Rightarrow \dfrac{1}{a}\left( {\dfrac{{ - r}}{{1 + r}}} \right) = \dfrac{1}{a}\left( {\dfrac{{ - {r^2}}}{{\left( {1 + r} \right)\left( {1 + r + {r^2}} \right)}}} \right)$
$\Rightarrow \dfrac{{ - r}}{{a\left( {1 + r} \right)}} = \dfrac{{ - {r^2}}}{{a\left( {1 + r + {r^2}} \right)}}$
After some arithmetic simplification, we get;
$\Rightarrow \dfrac{1}{{\left( {1 + r} \right)}} \ne \dfrac{{ - r}}{{\left( {1 + r + {r^2}} \right)}}$
Means; $\dfrac{1}{{{b_2}}} - \dfrac{1}{{{b_1}}} \ne \dfrac{1}{{{b_3}}} - \dfrac{1}{{{b_2}}}$
Therefore, the numbers ${b_1},{b_2},{b_3},{b_4}$ are not in H.P. , so statement II is not true.
So the conclusion is, statement I is true but statement II is false.
Hence the correct answer for this question is option $\left( 3 \right)$.

Note: Basic knowledge about properties and formulae of G.P. ,H.P., A.P. is very helpful for solving this
type of questions. $\left( 1 \right)$ Harmonic progression (H.P.) : Harmonic progression is a type of sequence which is obtained by taking the reciprocals of the A.P. , example: if $a,b,c,d$ is in A.P. then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}$ will be in H.P. The ${n^{th}}$ term of a H.P. is given by: $\dfrac{1}{{\left[ {a + \left( {n - 1} \right)d} \right]}}$ (reciprocal of ${n^{th}}$ term of A.P.) . Sum of $n$ terms of H.P. is given by: {S_n} = \dfrac{1}{d}\ln \left\\{ {\dfrac{{2a + \left( {2n - 1} \right)d}}{{2a - d}}} \right\\} . $\left( 2 \right)$ Geometric progression: The ${n^{th}}$ term of a G.P. is given by ${a_r} = a{r^{n - 1}}$ . The formula for sum of $n$ terms of a G.P. is given by: ${S_n} = a\left[ {\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}} \right]{\text{ if }}r \ne 1$ . $\left( 3 \right)$ Arithmetic progression: The ${n^{th}}$ term of a A.P. is given by ${a_n} = a + \left( {n - 1} \right)d$ . The formula for sum of $n$ terms of a A.P. is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ .