Question

if a,b,c,d are in gp prove that (a^n+b^n), (b^n+c^n), (c^n+d^n) are in gp​

Answer 1

Concept used :-

If we want to show that a, b, c are in GP, then we have to show that

$\boxed{ \red{ \tt \: \dfrac{b}{a} = \dfrac{c}{b} }}$

$\large\underline\purple{\bold{Solution :- }}$

Let common ratio be r.

Since a, b, c, d are in GP.

So,

$\begin{gathered}\begin{gathered}\bf Let -  \begin{cases} &\sf{a = a} \\ &\sf{b \: = ar}\\ &\sf{c \: = {ar}^{2} }\\ &\sf{d \: = {ar}^{3} } \end{cases}\end{gathered}\end{gathered}$

Consider

$\tt \: \dfrac{ {b}^{n} + {c}^{n} }{ {a}^{n} + {b}^{n} }$

On substituting the values of a, b, c we get

$= \: \tt \: \dfrac{ {(ar)}^{n} + { {(ar}^{2}) }^{n} }{ {a}^{n} + {(ar)}^{n} }$

$= \tt \: \dfrac{ {a}^{n} {r}^{n} + {a}^{n} {r}^{2n} }{ {a}^{n} + {a}^{n} {r}^{n} }$

$= \tt \: \dfrac{{a}^{n} {r}^{n}(1 + {r}^{n}) }{{a}^{n}(1 + {r}^{n})}$

$= \tt \: {r}^{n} - - - (i)$

Consider

$\tt \: \dfrac{{c}^{n} + {d}^{n}}{{b}^{n} + {c}^{n}}$

On substituting the values of b, c, d, we get

$= \tt \: \dfrac{ { {(ar}^{2}) }^{n} + { ({ar}^{3} )}^{n} }{ {(ar)}^{n} + { {(ar}^{2}) }^{n} }$

$= \tt \: \dfrac{{a}^{n} {r}^{2n} + {a}^{n} {r}^{3n}}{{a}^{n} {r}^{n} + {a}^{n} {r}^{2n}}$

$= \tt \: \dfrac{{a}^{n} {r}^{2n}(1 + {r}^{n})}{{a}^{n} {r}^{n}(1 + {r}^{n} )}$

$= \tt \: {r}^{n} - - - (ii)$

From (i) and (ii), we conclude that

$\tt \: \dfrac{ {b}^{n} + {c}^{n} }{ {a}^{n} + {b}^{n} } = \dfrac{{c}^{n} + {d}^{n}}{{b}^{n} + {c}^{n}}$

$\: \tt \: ({a}^{n} + {b}^{n}),({b}^{n} + {c}^{n}),({c}^{n} + {d}^{n}) \: are \: in \: GP$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$