The $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an $A.P.$ are $a, b, c,$ respectively. Show that $(q-r) a+(r-p) b+(p-q) c=0$
If the numbers $a,\;b,\;c,\;d,\;e$ form an $A.P.$, then the value of $a - 4b + 6c - 4d + e$ is
The sum of $24$ terms of the following series $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + .........$ is
Let ${S_1},{S_2},......,{S_{101}}$ be the consecutive terms of an $A.P$ . If $\frac{1}{{{S_1}{S_2}}} + \frac{1}{{{S_2}{S_3}}} + .... + \frac{1}{{{S_{100}}{S_{101}}}} = \frac{1}{6}$ and ${S_1} + {S_{101}} = 50$ , then $\left| {{S_1} - {S_{101}}} \right|$ is equal to
Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. If the sum of this AP is $189,$ then $a_{6} \mathrm{a}_{16}$ is equal to :