If a, b and c are in arithmetic progression and a^2, b^2 and c^2 are in harmonic progression, then prove that either a=b=c or a, b and -c / 2 are in geometric progression: Sequence & Series (Mathematics)

Given that $2 b=a+c$ $a^2, b^2, c^2$ are in H.P. and $\\b^2=\frac{2 a^2 c^2}{a^2+c^2}\\$ From (2) $b^2=\frac{2 a^2 c^2}{4 b^2-2 a c}\\$ using (1) $\Rightarrow\left(a c-b^2\right)\left(a c+2 b^2\right)=0 \\ \Rightarrow b^2=a c$ or $2 b^2=-a c\\$ Case I: $b^2=a c$ $\Rightarrow\left(\frac{a+c}{2}\right)^2=a c\\$ using (1) $\Rightarrow a=c \Rightarrow a=b=c\\$ as $a, b, c$ , are in A.P. $\\$ Case II : $2 b^2=-a c$ $\Rightarrow a, b,-c / 2$ are in G.P.

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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