Geometric series - positive and negative ratio

jackcr
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Hello,

Second term of a geometric series is 48 and the fourth term is 3... Show that one possible value for the common ratio, r, of the series is -1/4 and state the other value.

ar=48, ar^3= 3... so ar^3/ar=3/48 which simplifies to r^2 = 1/16, therefore r = 1/4

Can anyone explain where the other solution is from? Or where I am wrong

Thanks, and sorry if this is in the wrong section, I'm not familiar with pre/post calculus
 
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Oh, nevermind its because I have rooted the 1/16 meaning it could have been + or - 1/4 to start with. Haha
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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