If a_n = - 24n^2 - 16n + 15, then a_1 + a_2 + + a_25 is equal to

aligned& (b) : a_n=-24 n^2-16 n+15=-2(2 n-3)(2 n-5) \\& a_n=12 n-3-12 n-5aligned \\ For n=1, a_1=-1+13 ; n=2, a_2=1+1 ; n=3, \\ a_3=13-1 \\ aligned& n=4, a_4=15-13 ; n=5, a_5=17-

If a_{n} = \frac{- 2}{4n^{2} - 16n + 15}, then a_{1} + a_{2} + + a_{25} is equal to: Sequence & Series (Mathematics)

$\begin{aligned}& \text { (b) }: a_n=\frac{-2}{4 n^2-16 n+15}=\frac{-2}{(2 n-3)(2 n-5)} \\& a_n=\frac{1}{2 n-3}-\frac{1}{2 n-5}\end{aligned}$ $\\$ For $n=1, a_1=-1+\frac{1}{3} ; n=2, a_2=1+1 ; n=3, \\ a_3=\frac{1}{3}-1$ $\\ \begin{aligned}& n=4, a_4=\frac{1}{5}-\frac{1}{3} ; n=5, a_5=\frac{1}{7}-\frac{1}{5} \\& n=25, a_{25}=\frac{1}{47}-\frac{1}{45}\end{aligned} \\$ Now, $a_1+a_2+a_3+a_4+a_5+\ldots . a_{25}$ $\\ \begin{aligned}&=\left(-1+\frac{1}{3}\right)+(1+1)+\left(\frac{1}{3}-1\right)+ \\& \left(\frac{1}{5}-\frac{1}{3}\right)+\left(\frac{1}{7}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{47}-\frac{1}{45}\right) \\& =-1+\frac{1}{3}+2-1+\frac{1}{47}=\frac{50}{141}\end{aligned} \\$

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