Let 2^nd ,8^th andn 44^th terms of a non-constant A. P. ben respectively the 1^st ,2^nd andn 3^rd terms of a G. P. If the n first term of the A. P. is 1, then the sum of its first 20 terms isn equal to

n 1+ d, 1+7 d, 1+43 d are in GP n (1+7 d)^2=(1+ d)(1+43 d) n 1+49 d^2+14 d=1+44 d+43 d^2 n 6 d^2-30 d=0 n d=5 n S_20=(20 / 2)[2 x 1+(20-1) x 5] n =10[2+95] n =970nn n

Let 2^{{nd}{~}},8^{{th}{~}} and 44^{{th}{~}} terms o... | Sequence & Series (Mathematics)

\begin{aligned}\n& 1+\mathrm{d}, 1+7 \mathrm{~d}, 1+43 \mathrm{~d} \text { are in GP } \\\n& (1+7 \mathrm{~d})^2=(1+\mathrm{d})(1+43 \mathrm{~d}) \\\n& 1+49 \mathrm{~d}^2+14 \mathrm{~d}=1+44 \mathrm{~d}+43 \mathrm{~d}^2 \\\n& 6 \mathrm{~d}^2-30 \mathrm{~d}=0 \\\n& \mathrm{~d}=5 \\\n& \mathrm{~S}_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] \\\n& =10[2+95] \\\n& =970\n\n\end{aligned}\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