Question

If a,b,c,d are in GP prove that (a^n+b^n),(b^n+c^n) and (c^n+a^n) are in gp

Answer 1

Step-by-step explanation:

Given If a,b,c,d are in GP prove that (a^n+b^n),(b^n+c^n) and (c^n+a^n) are in gp

• We need to prove that (a^n + b^n), (b^n + c^n), (c^n + d^n) are in G.P
• We need to show b^n + c^n / a^n + b^n = c^n + d^n / b^n + c^n
• So b^n + c^n / a^n + b^n
• Put b = ar and c = ar^2 so we get
• = (ar)^n + (ar^2)^n / a^n + (ar)^n
• = a^nr^n + a^n r^2n / a^n + a^n r^n
• = a^n r^n (1 + r^n) / a^n (1 + r^n)
• = r^n
• Now putting c = ar^2, d = ar^3, b = ar on the right hand side we get
• = (ar^2)^n + (ar^3)^n / (ar)^n + (ar^2)^n
• = a^nr^2n + a^n.r^2n / a^nr^n + a^n r^2n
• = a^n(r^2n + r^3n) / a^n (r^n + r^2n)
• = (r^2n + r^3n) / (r^n + r^2n)
• = r^2n (1 + r) / r^n (1 + r)
• = r^2n / r^n
• = r^n . r^n / r^n
• =  r^n

Now both left hand side and right hand side are equal to r^n.. hence proved.

Reference link will be

https://brainly.in/question/1663277

Answer 2

$(a^{n} + b^{n})$, $(b^{n} + c^{n})$, and $(c^{n} + d^{n})$ are in G.P.

Step-by-step explanation:

Given that, a, b, c, and d are in G.P.

Let, b = ar, c = ar² and d = ar³, where r is the common ratio.

Now, we have to prove that $(a^{n} + b^{n})$, $(b^{n} + c^{n})$, and $(c^{n} + d^{n})$ are also in G.P.

Now, $\frac{b^{n} + c^{n}}{a^{n} + b^{n}} = \frac{(ar)^{n} + (ar^{2})^{n}}{a^{n} + (ar)^{n}} = r^{n}$

Again, $\frac{c^{n} + d^{n}}{b^{n} + c^{n}} = \frac{(ar^{2} )^{n} + (ar^{3})^{n}}{(ar)^{n} + (ar^{2} )^{n}} = r^{n}$

Therefore, we can conclude that $(a^{n} + b^{n})$, $(b^{n} + c^{n})$, and $(c^{n} + d^{n})$ are in G.P. (Answer)