The fourth power of the common difference of an arithmetic progressionn with integer entries is added to the product of any four consecutiven terms of it, prove that the resulting sum is square of an integer.

Let four consecutive terms of the A.P. be a-3 d, a-d, a+d and a+3 d. Common difference is 2 d.nGiven a-3 d, a-d, a+d and a+3 d are integers. Therefore, 2 d is also an integer.nNow, E=(a-3 d)(a-d)(a+d)(a+3 d)+(2 d)^4nnn n =(a^2-9 d^2)(a^2-d^2)+16 d^4 n =a^4-10 d^2 a^2+9 d^4+16 d^4n nnE=(a^2-5 d^2)^2 is an integern(As a-3 d, a+3 d and 2 d are integers => a^2-5 d^2 is also an integer)nThus, E is the square of an integer

The fourth power of the common difference of an arithme... | Sequence & Series (Mathematics)

Let four consecutive terms of the A.P. be

a-3 d, a-d, a+d

and

a+3 d

. Common difference is

2 d

.\nGiven

a-3 d, a-d, a+d

and

a+3 d

are integers. Therefore,

2 d

is also an integer.\nNow,

E=(a-3 d)(a-d)(a+d)(a+3 d)+(2 d)^4

\n\n

\n\begin{aligned}\n& =\left(a^2-9 d^2\right)\left(a^2-d^2\right)+16 d^4 \\\n& =a^4-10 d^2 a^2+9 d^4+16 d^4\n\end{aligned}\n

E=\left(a^2-5 d^2\right)^2

is an integer\n(As

a-3 d, a+3 d

and

2 d

are integers

\Rightarrow a^2-5 d^2

is also an integer)\nThus,

E

is the square of an integer

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

Download Divine JEE App

Quick Links

Exam Categories

Contact Info

Divine JEE