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Answers posted by rvidyagovindarajan_1

917
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answered Feb 24, 2014
Toolbox:$r^{th}A.M.$ between $a\:\;and\:\:b=A_r=a+rd$ where $d=\large\frac{b-a}{n+1}$$\:\:and\:\:n...
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answered Feb 24, 2014
Toolbox:The $A.M.$ between $a\:\;and\:\:b$ is $\large\frac{a+b}{2}$Given that $\large\frac{a^n+b...
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answered Feb 24, 2014
Toolbox:$r^{th}$ A.M. between $a\:\:and\:\:b$ = $A_r=a+rd$ where $d=\large\frac{b-a}{n+1}$ and $...
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answered Feb 23, 2014
Toolbox:$S_n=\large\frac{n}{2}$$[2a+(n-1)d]$In any series $S_n-S_{n-1}=t_n$Given: sum of $n$ terms $...
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answered Feb 23, 2014
Toolbox:$S_n=\large\frac{n}{2}$$[2a+(n-1)d]$Given: Sum of $m$ terms : Sum of $n$ terms of an $A.P=m...
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answered Feb 23, 2014
Toolbox:$S_n=\large\frac{n}{2}$$[2A+(n-1)d]$ where $A$ is first term and $d$ is common difference....
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answered Feb 23, 2014
Toolbox:Sum of first $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[2a+(n-1)d]$Given that the sum of ...
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answered Feb 22, 2014
Toolbox:Sum of $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[2a+(n-1)d]$Given that the ratio of s...
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answered Feb 22, 2014
Toolbox:Sum of $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[2a+(n-1)d]$$t_n=a+(n-1)d$Given: Sum of...
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answered Feb 22, 2014
Toolbox:Sum of $n$ terms of an $A.P.=\large\frac{n}{2}$$[l+a]$ where $a=$ first term and $l=$last ...
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answered Feb 22, 2014
Toolbox:Sum of $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[(2a+(n-1)d]$$n^{th}\:term=t_n=a+(n-1)d$...
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0 votes
answered Feb 20, 2014
Toolbox:Sum of $n$ terms of an A.P.$=S_n=\large\frac{n}{2}$$[2a+(n-1)d]$Given $A.P.$ is $-6,-\la...
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answered Feb 19, 2014
Toolbox:Sum of first $n$ terms of an $A.P.$=$S_n=\large\frac{n}{2}$$[2a+(n-1)d]$$t_n=a+(n-1)d$Given:...
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answered Feb 19, 2014
Toolbox:$n^{th}$ term of an A.P., $t_n=a+(n-1)d$ where $a=$First term and $d=$common differenceSum ...
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answered Feb 18, 2014
Toolbox:$n^{th}$ term of an $A.P.$ is $t_n=a+(n-1)d$ where $a$ is first term and $d$ is common...
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answered Feb 18, 2014
Given $a_n=a_{n-1}+a_{n-2}$ and $a_1=a_2=1$By putting $n=3,4,......$ we get$a_3=a_2+a_1=1+1=...
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answered Feb 18, 2014
Given: $a_1=a_2=2\:\:and\:\:a_n=a_{n-1}-1$By putting $n=3,4... $ we get the first 5 terms as$a_3=...
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answered Feb 18, 2014
Given: $a_1=-1\:\:and\:\:a_n=\large\frac{a_{n-1}}{n}$By putting $n=2,3,.......$ we get the first 5...
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answered Feb 18, 2014
Given: $a_1=3$ and $a_n=3a_{n-1}+2$By putting $n=2,3........... $ we get the first 5 terms as$...
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answered Feb 18, 2014
$a_n=\large\frac{n(n-2)}{n+3}$By putting $n=20$ we get the $20^{th}$ term as$a_{20}=\large\frac{2...
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answered Feb 18, 2014
Given: $a_n=(-1)^{n-1}.n^3$By putting $n=9$ we get the $9^{th}$ term as $a_9=(-1)^8.9^3=729$
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answered Feb 18, 2014
Given: $a_n=\large\frac{n^2}{2^n}$By putting $n=7$ we get the $7^{th}$ term as$a_7=\large\frac{7^...
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answered Feb 17, 2014
Given: General term of a sequence, $a_n=4n-3$By putting $n=17$ we get the $17^{th}$ term = $a_{...
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answered Feb 17, 2014
Given: $t_n=n.\large\frac{n^2+5}{4}$By putting $n=1,2,3...$ we get the first 5 terms as $t_1=1.\la...
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answered Feb 17, 2014
Given: $t_n=(-1)^{n-1}.\:5^{n+1}$By putting $n=1,2,3.....$ we get the first 5 terms as$t_1=(-1)^{1...
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answered Feb 17, 2014
Given: $t_n=\large\frac{2n-3}{6}$By putting $n=1,2,3.......$ we can get the first 5 terms as$t_1=\...
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answered Feb 17, 2014
Given: $t_n=2^n$By putting $n=1,2,3......$ we can get the first 5 terms as$t_1=2^1=2$$t_2=2^2=4$$t...
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answered Feb 17, 2014
Given: $t_n=\large\frac{n}{n+1}$By putting $n=1,2.....$ in $t_n$ we can get the first 5 terms as$t...
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answered Feb 15, 2014
Given: 6 boys and 6 girls.$\therefore\:$ There are 12 persons.All of them can be seated in $(12)!$...
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answered Feb 15, 2014
Toolbox:If $n^{th}$ term is known, put $n=1,2,3......$ to get the series.Given: $t_n=n(n+2)$$\t...
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answered Feb 12, 2014
Total probability theorem:If $E_1,E_2,E_3,.............E_n$ are mutually exclusive and exhaustiv...
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answered Feb 10, 2014
Toolbox:The area enclosed by a curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $y=b$ is given...
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answered Feb 10, 2014
Toolbox: Whenever a function is represented by y=|x| two cases arises. (i) y=x if $x\geq 0;$ ...
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answered Feb 6, 2014
Toolbox:If a matrix is singular then its determinant is zero.$cosec^{-1}x+sec^{-1}x=\large\frac{\pi}...
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answered Feb 5, 2014
Toolbox: $\sin x-\sin y=2\cos(\large\frac{x+y}{2})$$\sin(\large\frac{x-y}{2})$ $\cos...
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answered Feb 4, 2014
Toolbox:To obtain the absolute maxima or minima for the function $f(x)$(i) Find $f'(x)$ and put $ f'...
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answered Feb 4, 2014
Toolbox: If we are given two curves represented by y=f(x);y=g(x),where $f(x)\geq g(x)$ in [a,b],t...
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answered Feb 3, 2014
Toolbox: Direction ratios of a given vector is whose points are $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2...
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answered Feb 3, 2014
Toolbox:If two lines are $\perp$ then $a_1a_2+b_1b_2+c_1c_2=0$Where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_...
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answered Feb 3, 2014
Toolbox:$(\overrightarrow x-\overrightarrow a).(\overrightarrow x+\overrightarrow a)=|\overrightarro...
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answered Feb 3, 2014
Toolbox:$|\overrightarrow a+\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2+2\overr...
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answered Jan 31, 2014
Toolbox:Baye's Theorem:Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, ...
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answered Jan 31, 2014
Toolbox: A determinant of order $2\times 2$ can be evaluated as $\begin{vmatrix}a_{11} & a_{1...
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answered Jan 30, 2014
Toolbox:For a matrix $A$ which is non singular,$A^{-1}=\large\frac{1}{|A|}$$(Adj\:A)$$A.A^{-1}=I$Giv...
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answered Jan 30, 2014
Toolbox:$A^{-1}=\large\frac{1}{|A|}$$. (Adj \:A)$$A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix...
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answered Jan 28, 2014
Toolbox: \( tan^{-1}x+cot^{-1}x=\large\frac{\pi}{2} \) \(tan^{-1}1=\large\frac{\pi}{4}\) Ans...
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answered Jan 28, 2014
Toolbox:Take \( sin^{-1}\frac{3}{4}=x\) and proceed\( cosx=2cos^2\frac{x}{2}-1\)\(cosx=\sqrt{1-sin^2...
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