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

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

If a, b and c are in arithmetic progression and a^2, b^... | 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

Let a_{1},a_{2},a_{3},. be a G.P. of increasing positive numbers. Let the sum of its 6^{{th}{~}} and 8^{{th}{~}} terms be 2 and the product of its 3^{{rd}{~}} and 5^{{th}{~}} terms be 1/9. Then 6( a_{2} + a_{4} )( a_{4} + a_{6} ) is equal to

For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + ( - 1)^{n - 1}1^{3} =

The sum of the first n terms of the series (1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2 . 6^{2} + ..) is (\frac{n(n + 1)^{2}}{2}, when n is even. when n is odd, the sum is

Let a,b be two non-zero real numbers. If p and r are the roots of the equation x^{2} - 8ax + 2a = 0 and q and s are the roots of the equation \end{enumerate} x^{2} + 12bx + 6b = 0, such that \frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s} are in A.P., then a^{- 1} - b^{- 1} is equal to

If a_{1},a_{2},a_{3}, are in A.P. such that a_{1} + a_{7} + a_{16} = 40, then the sum of the first 15 terms of this A.P. is

Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

Number of 4 -digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

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