If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of last and first term is $273$, then the numbers are

If $a,\;b,\;c$ are in $A.P.$, then $\frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} = $

Jairam purchased a house in Rs. $15000$ and paid Rs. $5000$ at once. Rest money he promised to pay in annual installment of Rs. $1000$ with $10\%$ per annum interest. How much money is to be paid by Jairam $\mathrm{Rs.}$ ...................

Given that $n$ A.M.'s are inserted between two sets of numbers $a,\;2b$and $2a,\;b$, where $a,\;b \in R$. Suppose further that ${m^{th}}$ mean between these sets of numbers is same, then the ratio $a:b$ equals

Let $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},.....,$  $({x_i} \ne \,0\,for\,\,i\, = 1,2,....,n)$  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\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}}} \right)} $  is equal to.

Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an $A.P.$ If $d$ is the common difference of this $A.P.$, then $50-\frac{2 d}{\beta^{2}}$ is equal to.