
If a, b, c, d are in G.P. , prove that $\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$ are in G.P.
Answer
597.6k+ views
Hint – We will start solving this question by collecting all the information mentioned in the question and by using this information we will find some values and make some equations to prove that $\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$are in G.P, i.e., Geometric Progression.
Complete step-by-step answer:
Now, it is mentioned in the question that a, b, c, d are in G.P.
Therefore, we can write,
$\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}$
Now,
$\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c} = k$(say)
$ \Rightarrow \dfrac{b}{a} = k,\dfrac{c}{b} = k,\dfrac{d}{c} = k$
$
\Rightarrow b = ak,c = bk,d = ck \\
\Rightarrow b = ak,c = ak \times k,d = ak \times k \times k \\
$
$ \Rightarrow b = ak,c = a{k^2},d = a{k^3}$ ………… (1)
Now, $\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$ will be in G.P,
if $\dfrac{{{b^n} + {c^n}}}{{{a^n} + {b^n}}} = \dfrac{{{c^n} + {d^n}}}{{{b^n} + {c^n}}}$
if ${\left( {{b^n} + {c^n}} \right)^2} = \left( {{a^n} + {b^n}} \right)\left( {{c^n} + {d^n}} \right)$
Now, taking LHS of this equation,
${\left( {{b^n} + {c^n}} \right)^2}$
$ = \left[ {{{\left( {ak} \right)}^n} + {{\left( {a{k^2}} \right)}^n}} \right]$ [ Using (1) ]
$
= {\left( {{a^n}{k^n} + {a^n}{k^{2n}}} \right)^2} \\
= {\left[ {{a^n}{k^n}\left( {1 + {k^n}} \right)} \right]^2} \\
= {a^{2n}}{k^{2n}}{\left( {1 + {k^n}} \right)^2} \\
$
Now, taking RHS of the above equation,
$\left( {{a^n} + {b^n}} \right)\left( {{c^n} + {d^n}} \right)$
$ = \left( {{a^n} + {a^n}{k^n}} \right)\left( {{a^n}{k^{2n}} + {a^n}{k^{3n}}} \right)$ [ Using (1) ]
$
= {a^n}\left( {1 + {k^n}} \right)\left[ {{a^n}{k^{2n}}\left( {1 + {k^n}} \right)} \right] \\
= {a^{2n}}{k^{2n}}{\left( {1 + {k^n}} \right)^2} \\
$
$\therefore $ LHS = RHS
Hence, $\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$ are in G.P.
Note – A Geometric Progression, i.e., G.P, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. These kinds of questions must be solved with concentration because this is the only method to solve this question.
Complete step-by-step answer:
Now, it is mentioned in the question that a, b, c, d are in G.P.
Therefore, we can write,
$\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}$
Now,
$\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c} = k$(say)
$ \Rightarrow \dfrac{b}{a} = k,\dfrac{c}{b} = k,\dfrac{d}{c} = k$
$
\Rightarrow b = ak,c = bk,d = ck \\
\Rightarrow b = ak,c = ak \times k,d = ak \times k \times k \\
$
$ \Rightarrow b = ak,c = a{k^2},d = a{k^3}$ ………… (1)
Now, $\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$ will be in G.P,
if $\dfrac{{{b^n} + {c^n}}}{{{a^n} + {b^n}}} = \dfrac{{{c^n} + {d^n}}}{{{b^n} + {c^n}}}$
if ${\left( {{b^n} + {c^n}} \right)^2} = \left( {{a^n} + {b^n}} \right)\left( {{c^n} + {d^n}} \right)$
Now, taking LHS of this equation,
${\left( {{b^n} + {c^n}} \right)^2}$
$ = \left[ {{{\left( {ak} \right)}^n} + {{\left( {a{k^2}} \right)}^n}} \right]$ [ Using (1) ]
$
= {\left( {{a^n}{k^n} + {a^n}{k^{2n}}} \right)^2} \\
= {\left[ {{a^n}{k^n}\left( {1 + {k^n}} \right)} \right]^2} \\
= {a^{2n}}{k^{2n}}{\left( {1 + {k^n}} \right)^2} \\
$
Now, taking RHS of the above equation,
$\left( {{a^n} + {b^n}} \right)\left( {{c^n} + {d^n}} \right)$
$ = \left( {{a^n} + {a^n}{k^n}} \right)\left( {{a^n}{k^{2n}} + {a^n}{k^{3n}}} \right)$ [ Using (1) ]
$
= {a^n}\left( {1 + {k^n}} \right)\left[ {{a^n}{k^{2n}}\left( {1 + {k^n}} \right)} \right] \\
= {a^{2n}}{k^{2n}}{\left( {1 + {k^n}} \right)^2} \\
$
$\therefore $ LHS = RHS
Hence, $\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$ are in G.P.
Note – A Geometric Progression, i.e., G.P, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. These kinds of questions must be solved with concentration because this is the only method to solve this question.
Recently Updated Pages
Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 6 Maths: Engaging Questions & Answers for Success

Class 6 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

