Let a_{1},a_{2},a_{3}, be a sequence of positive integers in arithmetic progression with common difference 2 . Also, let b_{1},b_{2},b_{3}, be a sequence of positive integers in geometric progression with common ratio 2 . If a_{1} = b_{1} = c, then the number of all possible values of c, for which the equality 2( a_{1} + a_{2} + + a_{n} ) = b_{1} + b_{2} + + b_{n} holds for some positive integer n, is

Question

Let a_1,a_2,a_3, ... be a sequence of positive integers in arithmetic progression with common difference 2 . Also, let b_1,b_2,b_3, ... be a sequence of positive integers in geometric progression with common ratio 2 . If a_1 = b_1 = c, then the number of all possible values of c, for which the equality 2( a_1 + a_2 + ... + a_n ) = b_1 + b_2 + ... + b_n holds for some positive integer n, is \_\_\_\_

DivineJEE · Accepted Answer

\ (1) : 2 sum _i=1^n a_i=sum _i=1^n b_i => 2 * (n / 2)(2 c+(n-1) * 2)=c((2^n-1 / 2-1)) => n(2 c+2 n-2)=c(2^n-1) => c(2^n-2 n-1)=n(2 n-2) => c=(2 n(n-1) / 2^n-2 n-1) >= 1\ [As, c must be positive integer and n >= 2 as for n=1, c=0, not possible] \ => 2 n(n-1) >= 2^n-2 n-1 => 2^n <= 2 n^2-2 n+2 n+1 => 2^n <= 2 n^2+1\ The above inequality holds for n <= 6. Taking n=2,3,4,5,6 using (i), we get For n=2, c=(2 * 2 * 1 / 4-4-1)=- ve, discarded \ For n=3, c=(4-4-3-1 / 8-6-1)=12 \ For n=4, c=frac(8-6-1 / 2 * 4 * 3)16-9-1=(24 / 7), not an integer \ For n=5, c=frac(16 * 5 * 4 / 2 * 5 * 10-1)32-(40 / 21), not an integer \ For n=6, c=(2 * 6 * 5 / 64-12-1)=(60 / 51), not an integer \ Thus, c=12 is the only possible value. Hence, the number of values of c is 1.

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