Solveeit Logo
Login

Question

Question: If a, b, c, d are in G.P. , prove that \(\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \righ...

If a, b, c, d are in G.P. , prove that (an+bn),(bn+cn),(cn+dn)\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right) are in G.P.

Explanation

Solution

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 (an+bn),(bn+cn),(cn+dn)\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,
ba=cb=dc\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}

Now,
ba=cb=dc=k\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c} = k(say)
⇒ba=k,cb=k,dc=k\Rightarrow \dfrac{b}{a} = k,\dfrac{c}{b} = k,\dfrac{d}{c} = k
⇒b=ak,c=bk,d=ck ⇒b=ak,c=ak×k,d=ak×k×k  \Rightarrow b = ak,c = bk,d = ck \\\ \Rightarrow b = ak,c = ak \times k,d = ak \times k \times k \\\
⇒b=ak,c=ak2,d=ak3\Rightarrow b = ak,c = a{k^2},d = a{k^3} ………… (1)

Now, (an+bn),(bn+cn),(cn+dn)\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right) will be in G.P,
if bn+cnan+bn=cn+dnbn+cn\dfrac{{{b^n} + {c^n}}}{{{a^n} + {b^n}}} = \dfrac{{{c^n} + {d^n}}}{{{b^n} + {c^n}}}
if (bn+cn)2=(an+bn)(cn+dn){\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,
(bn+cn)2{\left( {{b^n} + {c^n}} \right)^2}
=[(ak)n+(ak2)n]= \left[ {{{\left( {ak} \right)}^n} + {{\left( {a{k^2}} \right)}^n}} \right] [ Using (1) ]
=(ankn+ank2n)2 =[ankn(1+kn)]2 =a2nk2n(1+kn)2  = {\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,
(an+bn)(cn+dn)\left( {{a^n} + {b^n}} \right)\left( {{c^n} + {d^n}} \right)
=(an+ankn)(ank2n+ank3n)= \left( {{a^n} + {a^n}{k^n}} \right)\left( {{a^n}{k^{2n}} + {a^n}{k^{3n}}} \right) [ Using (1) ]
=an(1+kn)[ank2n(1+kn)] =a2nk2n(1+kn)2  = {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, (an+bn),(bn+cn),(cn+dn)\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.