Question

If a:b = b:c = c:d, prove that (a+b) = (a2+b2+c2) : (b2+c2+d2)​

Answer 1

Please reframe your question, as it is unclear what you meant.

Anyway, this how I saw your question:

If a:b = b:c = c:d, prove that a+b = (a^2+b^2+c^2) : (b^2+c^2+d^2)​

Solution

Let

$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{1}{k} \:\:\:[k\:is\:a\:constant]$

$\\\implies\frac{a}{b}=\frac{1}{k} \\\implies\frac{b}{a}=k\\ \implies b=ak \:--\:(1)$

Also,

$\\\implies\frac{b}{c}=\frac{1}{k} \\\implies\frac{c }{b}=k\\ \implies c=bk=(ak)k\:--\:[from\:(1)]\\\implies c = ak^2\:--\:(2)$

And,

$\\\implies\frac{c}{d}=\frac{1}{k} \\\implies\frac{d}{c}=k\\ \implies d=ck=(ak^2)k\:--\:[from\:(2)]\\\implies d = ak^3\:--\:(3)$

Now,

$\frac{a^2+b^2+c^2}{b^2+c^2+d^2} =\frac{a^2+(ak)^2+(ak^2)^2}{(ak)^2+(ak^2)^2++(ak^3)^2}\\=\frac{a^2+a^2k^2+a^2k^4}{a^2k^2+a^2k^4+a^2k^6}=\frac{a^2(1+k^2+k^4)}{a^2(k^2+k^4+k^6)}\\=\frac{1+k^2+k^4}{k^2+k^4+k^6}=\frac{1+k^2+k^4}{k^2(1+k^2+k^4)}\\=\frac{1}{k^2}=(\frac{a}{b})^2$

But, $\frac{a^2+b^2+c^2}{b^2+c^2+d^2}\neq a+b$

Maybe you meant

"If a:b = b:c = c:d, prove that (a/b)^2 = (a^2+b^2+c^2) : (b^2+c^2+d^2)"?

I don't know