The 8^th common term of the series S_1 = 3 + 7 + 11 + 15 + 19 + ... .. S_2 = 1 + 6 + 11 + 16 + 21 + ... .. , is

S_1 = 3 + 7 + 11 + 15 + 19 + ... . a_1 = 3,d_1 = 4 \ S_2 = 1 + 6 + 11 + 16 + 21 + ... \ a_2 = 1,d_2 = 5 \ First common term, a = 11 Common difference, d = LCM( d_1,d_2 ) = LCM(4,5) = 20 8^th common term . of the series a_8 = a + 7d = 11 + 7 x 20 = 151

The 8^{{th}{~}} common term of the series S_{1} = 3... | Sequence & Series (Mathematics)

$S_{1} = 3 + 7 + 11 + 15 + 19 + \ldots.\\$ $\begin{matrix} & a_{1} = 3,d_{1} = 4 \\ & S_{2} = 1 + 6 + 11 + 16 + 21 + \ldots \\ & a_{2} = 1,d_{2} = 5 \end{matrix}\\$ First common term, $a = 11\\$ Common difference, $d = LCM\left( d_{1},d_{2} \right) \\= LCM(4,5) = 20\\$ $8^{\text{th}\text{~}}$ common term . of the series $a_{8} = a + 7d \\= 11 + 7 \times 20 = 151$

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