Let the digits a,b,c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?: Sequence & Series (Mathematics)

Let the digits

a,b,c

be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

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

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?

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

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

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}

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

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}}

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