Let a_1, a_2, a_3, s be an A.P. If sum _r=1^infinity fraca_r2^r = 4, then 4a_2 is equal to

a_1, a_2, ... a_n, ... be an A.P. and sum _r=1^infinity (a_r / 2^r)=4 Let S=sum _r=1^infinity (a_r / 2^r)=(a_1 / 2)+(a_2 / 2^2)+(a_3 / 2^3)+ * s \ (s / 2)=(a_1 / 2^2)+(a_2 / 2^3)+(a_3 / 2^4)+ * s S-(s / 2)=(a_1 / 2)+(a_2-a_1 / 2^2)+(a_3-a_2 / 2^3)+ * s => (S / 2)=(a_1 / 2)+d((1 / 2^2)+(1 / 2^3)+ * s)=\ (a_1 / 2)+d(frac(1 / 4)1-(1 / 2))=(a_1 / 2)+(d / 2) => S=a_1+d=a_2=4 Given => 4 a_2=4 x 4=16

Let a_{1},a_{2},a_{3}, be an A.P. If \sum_{r = 1}^{\i... | Sequence & Series (Mathematics)

$a_1, a_2, \ldots a_n, \ldots$ be an A.P. and $\sum_{r=1}^{\infty} \frac{a_r}{2^r}=4\\$ Let $S=\sum_{r=1}^{\infty} \frac{a_r}{2^r}=\frac{a_1}{2}+\frac{a_2}{2^2}+\frac{a_3}{2^3}+\cdots$ $\\ \frac{s}{2}=\frac{a_1}{2^2}+\frac{a_2}{2^3}+\frac{a_3}{2^4}+\cdots \\$ $S-\frac{s}{2}=\frac{a_1}{2}+\frac{a_2-a_1}{2^2}+\frac{a_3-a_2}{2^3}+\cdots \\$ $\Rightarrow \frac{S}{2}=\frac{a_1}{2}+d\left(\frac{1}{2^2}+\frac{1}{2^3}+\cdots\right)$ = $\\ \frac{a_1}{2}+d\left(\frac{\frac{1}{4}}{1-\frac{1}{2}}\right)=\frac{a_1}{2}+\frac{d}{2} \\$ $\Rightarrow S=a_1+d=a_2=4 \\$ Given $\Rightarrow 4 a_2=4 \times 4=16$

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

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