If the 10^th term of an A.P. is (1 / 20)n and its 20^th term is (1 / 10), then then sum of its first 200 terms is

(d) : Given, T_10=(1 / 20)=a+9 d and T_20=(1 / 10)=a+19 dnSolving (i) and (ii), we have a=(1 / 200), d=(1 / 200)nnTherefore S_200=(200 / 2)[(2 / 200)+(199 / 200)]=(201 / 2)=100 (1 / 2)n

If the 10^{{th}{~}} term of an A.P. is \frac{1}{20} a... | Sequence & Series (Mathematics)

(d) : Given,

T_{10}=\frac{1}{20}=a+9 d

and

T_{20}=\frac{1}{10}=a+19 d

\nSolving (i) and (ii), we have

a=\frac{1}{200}, d=\frac{1}{200}

\n\therefore S_{200}=\frac{200}{2}\left[\frac{2}{200}+\frac{199}{200}\right]=\frac{201}{2}=100 \frac{1}{2}\n

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