\( 515 \)
\( 510 \)
In an arithmetic progression, the sum of the first \( n \) terms is given by the formula: \[ S_n = \frac{n}{2} (2a + (n - 1) d), \] where \( a \) is the first term and \( d \) is the common difference.
We are given \( S_{40} = 1030 \) and \( S_{12} = 57 \). From these, we can solve for \( a \) and \( d \). Then, we calculate \( S_{30} - S_{10} \) using the same formula.
Final Answer: \( S_{30} - S_{10} = 515 \).
Step 1: Recall the formulas for the sum of an arithmetic progression (AP).
The sum of first \( n \) terms of an AP is:
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \] where \( a \) is the first term and \( d \) is the common difference.
Step 2: Use the given information to formulate equations.
Given:
\[ S_{40} = 1030 \quad \Rightarrow \quad 1030 = \frac{40}{2}[2a + 39d] = 20(2a + 39d) \] \[ S_{12} = 57 \quad \Rightarrow \quad 57 = \frac{12}{2}[2a + 11d] = 6(2a + 11d) \]
From these, express:
\[ 2a + 39d = \frac{1030}{20} = 51.5 \] \[ 2a + 11d = \frac{57}{6} = 9.5 \]
Step 3: Solve the system of equations to find \( a \) and \( d \).
Subtract the second equation from the first:
\[ (2a + 39d) - (2a + 11d) = 51.5 - 9.5 \] \[ 28d = 42 \quad \Rightarrow \quad d = \frac{42}{28} = 1.5 \] Substitute \( d = 1.5 \) into second equation:
\[ 2a + 11 \times 1.5 = 9.5 \quad \Rightarrow \quad 2a + 16.5 = 9.5 \] \[ 2a = 9.5 - 16.5 = -7 \quad \Rightarrow \quad a = -3.5 \]
Step 4: Calculate \( S_{30} \) and \( S_{10} \).
\[ S_{30} = \frac{30}{2} [2a + 29d] = 15 (2a + 29d) \] Substitute \( a = -3.5 \), \( d = 1.5 \):
\[ 2a + 29d = 2 \times (-3.5) + 29 \times 1.5 = -7 + 43.5 = 36.5 \] \[ S_{30} = 15 \times 36.5 = 547.5 \] Similarly, \[ S_{10} = \frac{10}{2} [2a + 9d] = 5 (2a + 9d) \] \[ 2a + 9d = -7 + 13.5 = 6.5 \] \[ S_{10} = 5 \times 6.5 = 32.5 \]
Step 5: Calculate \( S_{30} - S_{10} \).
\[ S_{30} - S_{10} = 547.5 - 32.5 = 515 \] Given answer is 510, so verify calculation of \( S_{40} \) and \( S_{12} \).
Check initial sums:
\[ S_{40} = 20(2a + 39d) = 20(-7 + 58.5) = 20 \times 51.5 = 1030 \] \[ S_{12} = 6(2a + 11d) = 6(-7 + 16.5) = 6 \times 9.5 = 57 \] Values of \( a \) and \( d \) are correct.
Recalculate \( S_{30} \) and \( S_{10} \) for precision:
\[ S_{30} = 15 (2a + 29d) = 15(-7 + 43.5) = 15 \times 36.5 = 547.5 \] \[ S_{10} = 5 (2a + 9d) = 5(-7 + 13.5) = 5 \times 6.5 = 32.5 \] Difference:
\[ 547.5 - 32.5 = 515 \]
Final answer:
\[ \boxed{515} \]
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

Method used for separation of mixture of products (B and C) obtained in the following reaction is: 