Let the positive numbers a,b,c,d be in A.P. then abc,n abd,acd,bcd, are

(d) : As a, b, c, d are in A.P.n => d, c, b, a are in A.P. => (1 / d), (1 / c), (1 / b), (1 / a) in H.P.n => (a b c d / d), (a b c d / c), (a b c d / b), (a b c d / a) are in H.Pn => a b c, a b d, a c d, b c d are in H.P

Let the positive numbers a,b,c,d be in A.P. then abc, ... | Sequence & Series (Mathematics)

(d) : As

a, b, c, d

are in A.P.\n

\Rightarrow d, c, b, a

are in A.P.

\Rightarrow \frac{1}{d}, \frac{1}{c}, \frac{1}{b}, \frac{1}{a}

in H.P.\n

\Rightarrow \frac{a b c d}{d}, \frac{a b c d}{c}, \frac{a b c d}{b}, \frac{a b c d}{a}

are in H.P\n

\Rightarrow a b c, a b d, a c d, b c d

are in H.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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