If a,b,c are in A.P.

Question

If $a,b,c$ are in A.P. and $a^2,b^2,c^2$ are in G.P. such that $a<b<c$ and $a+b+c=\dfrac34$, then the value of $a$ is:

Accepted Answer

$a+c=2b$, and $a+b+c=	frac34 \Rightarrow 3b=	frac34 \Rightarrow b=	frac14$.

$b^4=a^2c^2 \Rightarrow ac=\pm\dfrac{1}{16}$. With $a+c=2b=\dfrac12$: $16a^2-8a\pm1=0$.

Taking the case giving $a<b$: $16a^2-8a-1=0 \Rightarrow a = \dfrac{8\pm\sqrt{128}}{32} = \dfrac14\pm\dfrac{1}{2\sqrt2}$.

Since $a<b=\dfrac14$: $a = \dfrac14-\dfrac{1}{2\sqrt2}$.

Solutions · 1