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
629.1k+ 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 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

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

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

