The 20^{{th}{~}} term from the end of the progression 20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},, - 129\frac{1}{4} is :: Sequence & Series (Mathematics) Question | DivineJEE

The 20^{{th}{~}} term from the end of the progression 20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},, - 129\frac{1}{4} is :: Sequence & Series (Mathematics) Question | DivineJEE | DivineJEE

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

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 \log_{3}2,\log_{3}( 2^{x} - 5 ) and \log_{3}( 2^{x} - \frac{7}{2} ) are in arithmetic progression, determine the value of x

The sum of all those terms, of the arithmetic progression 3, 8, 13,........373, which are not divisible by 3 , is equal to

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 \log_{3}2,\log_{3}( 2^{x} - 5 ),\log_{3}( 2^{x} - \frac{7}{2} ) are in an arithmetic progression, then the value of x is equal to

The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24 , then what is the length of its smallest side?

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

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

The real number x_{1},x_{2},x_{3} satisfying the equation x^{3} - x^{2} + \beta x + \gamma = 0 are in A.P. Find the intervals in which \beta and \gamma lie.

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