Denote by \(s_k\) the sum of 12 terms of the \(k\)-th arithmetic progression, whose first term is \(k\) and common difference is \(2k-1\). We recall that the sum of 12 terms of an A.P. with first term \(a\) and common difference \(d\) is \[ S_{12} = \frac{12}{2} [2a + (12 - 1)d] = 6(2a + 11d). \] Hence for the \(k\)-th progression, \[ s_k = 6(2k + 11(2k - 1)) = 6(2k + 22k - 11) = 6(24k - 11) = 144k - 66. \] We wish to find \[ \sum_{k=1}^{10} s_k = \sum_{k=1}^{10} (144k - 66) = 144 \sum_{k=1}^{10} k - 66 \sum_{k=1}^{10} 1. \] Recall \[ \sum_{k=1}^{10} k = \frac{10 \cdot 11}{2} = 55, \] and \[ \sum_{k=1}^{10} 1 = 10. \] Thus \[ \sum_{k=1}^{10} s_k = 144 \cdot 55 - 66 \cdot 10 = 7920 - 660 = 7260. \] Hence, \[ \boxed{ \sum_{k=1}^{10} s_k = 7260. } \]
The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is
The statement
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to
Let α, β(α > β) be the roots of the quadratic equation x2 – x – 4 = 0.
If \(P_n=α^n–β^n, n∈N\) then \(\frac{P_{15}P_{16}–P_{14}P_{16}–P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}\)
is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,