Question

If a, b, c are in G.P., then
A. ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$ are in G.P.
B. ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),}}{{\text{b}}^{\text{2}}}{\text{(a + c)}}$ are in G.P.
C. $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{c + a}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$ are in G.P.
D. None of the above

Answer 1

We’ll first write the conditional equation for a, b, c, and then will note the value of b and c in terms of a and common ratio to create a relation in a, b, c and then will find the required answer with help of those equations.

Complete step by step answer:

Given data: a, b, c are in G.P.
From the given data i.e. a, b, c are in G.P., we can say that the common ratio will remain the same
$\dfrac{{\text{b}}}{{\text{a}}}{\text{ = }}\dfrac{{\text{c}}}{{\text{b}}}..............{\text{(i)}} \\\ \Rightarrow {{\text{b}}^{\text{2}}}{\text{ = ac}}.........{\text{(ii)}} \\\$
Let the common ratio be ‘r’
$\therefore {\text{b = ar}}$
Squaring both sides of the above equation
${{\text{b}}^{\text{2}}}{\text{ = }}{\left( {{\text{ar}}} \right)^{\text{2}}} \\\ \Rightarrow {{\text{b}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{{\text{r}}^{\text{2}}} \\\$
Also,${\text{c = a}}{{\text{r}}^{\text{2}}}$
Squaring both sides of the above equation
${{\text{c}}^{\text{2}}}{\text{ = }}{\left( {{\text{a}}{{\text{r}}^{\text{2}}}} \right)^{\text{2}}} \\\ \Rightarrow {{\text{c}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{{\text{r}}^{\text{4}}} \\\$
From the value of ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$ we can see that they are in G.P. with a common ratio of ${\text{'}}{{\text{r}}^{\text{2}}}{\text{'}}$
Common ratio=$\dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{c}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}{\text{ = }}{{\text{r}}^{\text{2}}}$
Therefore, option(A) is correct

Note: We can also verify our solution with the help of an example let say 2,4,8 where
a=2
b=4
c=8
for option(A) ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$=${\text{4,16,64}}$
since $\dfrac{{{\text{16}}}}{{\text{4}}}{\text{ = }}\dfrac{{{\text{64}}}}{{{\text{16}}}}{\text{ = 4}}$
therefore, ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$are in G.P.
(A)${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$ are in G.P.
Therefore option (A) is correct
for option(B) ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),}}{{\text{b}}^{\text{2}}}{\text{(a + c)}}$=${{\text{2}}^{\text{2}}}{\text{(4 + 8),}}{{\text{8}}^{\text{2}}}{\text{(2 + 4),}}{{\text{4}}^{\text{2}}}{\text{(2 + 8)}}$
${\text{48,386,160}}$
since $\dfrac{{{\text{386}}}}{{{\text{48}}}}{\text{ = }}\dfrac{{{\text{193}}}}{{{\text{24}}}} \ne \dfrac{{{\text{160}}}}{{{\text{386}}}}{\text{ = }}\dfrac{{{\text{80}}}}{{{\text{193}}}}$
therefore, ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),}}{{\text{b}}^{\text{2}}}{\text{(a + c)}}$are not in G.P.
for option(C) $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{c + a}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$=$\dfrac{{\text{2}}}{{{\text{4 + 8}}}}{\text{,}}\dfrac{{\text{4}}}{{{\text{8 + 2}}}}{\text{,}}\dfrac{{\text{8}}}{{{\text{2 + 4}}}}$
$\dfrac{{\text{1}}}{{\text{6}}}{\text{,}}\dfrac{{\text{2}}}{{\text{5}}}{\text{,}}\dfrac{{\text{4}}}{{\text{3}}}$
since $\dfrac{{\dfrac{{\text{2}}}{{\text{5}}}}}{{\dfrac{{\text{1}}}{{\text{6}}}}}{\text{ = }}\dfrac{{{\text{12}}}}{{\text{5}}} \ne \dfrac{{\dfrac{{\text{4}}}{{\text{3}}}}}{{\dfrac{{\text{2}}}{{\text{5}}}}}{\text{ = }}\dfrac{{{\text{10}}}}{{\text{3}}}$
therefore, $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{c + a}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$are not in G.P.