For any three positive real numbers and , . Then...

aligned& (a): 9(25 a^2+b^2)+25(c^2-3 a c)=15 b(3 a+c) \\& (15 a)^2+(3 b)^2+(5 c)^2-(15 a)(5 c)-(15 a)(3 b)-(3 b) \\& (5 c)=0 \\& 12[(15 a-3 b)^2+(3 b-5 c)^2+(5 c-15 a)^2]=0 \\& (15 a-3 b)^2=0,(3 b-5 c)^2=0,(5 c-15 a)^2=0 \\& 15 a=3 b=5 c a1=b5=c3 b, c, a are in A.P. aligned...

For any three positive real numbers a,b and c, 9( 25a^{2} + b^{2} ) + 25( c^{2} - 3ac ) = 15b(3a + c). Then: Sequence & Series (Mathematics)

For any three positive real numbers

a,b

and

c

,\n

9\left( 25a^{2} + b^{2} \right) + 25\left( c^{2} - 3ac \right) = 15b(3a + c)

.\n Then

b,c

and

a

are in A.P.

a,b

and

c

are in A.P.

a,b

and

c

are in G.P.

b,c

and

a

are in G.P.

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

Different A.P. are constructed with the first term 100 , the last term 199, and integral common differences. The sum of the common differences of all such A.P. having at least 3 terms and at most 33 terms is

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

If a_1, a_2, a_n are in A.P. where a_i>0 for all i show that &\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}&=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}

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

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

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

Let A_{1},G_{1} and H_{1} denote the arithmetic, geometric and harmonic means . respectively of two distinct positive numbers. For n ? 2 let A_{n - 1} and H_{n - 1} have . arithmetic, geometric and harmonic means as A_{n},G_{n},H_{n} respectively. Which of the following statements is correct?

If a_{n} = \frac{3}{4} - ( \frac{3}{4} )^{2} + ( \frac{3}{4} )^{3} + ( - 1)^{n - 1}( \frac{3}{4} )^{n} and b_{n} = 1 - a_{n}, then find the least natural number n_{0} such that b_{n} > a_{n}\forall n ? n_{0}

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