Question

The roots of the quadratic equation \( 3x^2 - px + q = 0 \) are the 10th and 11th terms of an arithmetic progression with common difference \( \frac{3}{2} \). If the sum of the first 11 terms of this arithmetic progression is 88, then \( q - 2q \) is equal to:

Hint:

In problems involving arithmetic progressions, use the sum and product of the roots formula for quadratic equations to find the required values.

Answer 1

Correct Answer: 474

Step 1: Given information

The sum of the first 11 terms of an arithmetic progression (A.P.) is given as: \[ S_{11} = \frac{11}{2}(2a + 10d) = 88 \]

Simplifying this, we get:

\[ a + 5d = 8 \]

Step 2: Finding the first term \( a \)

We are also given that the common difference \( d = \frac{3}{2} \). Substitute this value in the above equation: \[ a + 5 \times \frac{3}{2} = 8 \] \[ a + \frac{15}{2} = 8 \] \[ a = 8 - \frac{15}{2} = \frac{1}{2} \]

Step 3: Finding the 10th and 11th terms

The formula for the \( n^{th} \) term of an A.P. is: \[ T_n = a + (n-1)d \] Therefore, \[ T_{10} = a + 9d = \frac{1}{2} + 9 \times \frac{3}{2} = \frac{1}{2} + \frac{27}{2} = 14 \] \[ T_{11} = a + 10d = \frac{1}{2} + 10 \times \frac{3}{2} = \frac{1}{2} + 15 = \frac{31}{2} \]

Step 4: Finding the value of \( p \)

We are given: \[ \frac{p}{3} = T_{10} + T_{11} \] Substitute the values of \( T_{10} \) and \( T_{11} \): \[ \frac{p}{3} = 14 + \frac{31}{2} = \frac{28 + 31}{2} = \frac{59}{2} \] \[ p = \frac{3 \times 59}{2} = \frac{177}{2} \]

Step 5: Finding the value of \( q \)

\[ \frac{q}{3} = T_{10} \times T_{11} \] Substitute the values: \[ \frac{q}{3} = 14 \times \frac{31}{2} = 7 \times 31 = 217 \] \[ q = 3 \times 217 = 651 \]

Step 6: Finding the final expression \( q - 2p \)

\[ q - 2p = 651 - 2 \times \frac{177}{2} \] \[ q - 2p = 651 - 177 = 474 \]

Final Answer:

\[ \boxed{q - 2p = 474} \]

Answer 2

The \( 10^{\text{th}} \) and \( 11^{\text{th}} \) terms of the arithmetic progression are: \[ T_{10} = a + 9d \quad \text{and} \quad T_{11} = a + 10d, \] where \( a \) is the first term, and \( d = \frac{3}{2} \) is the common difference. We are also given that the sum of the first 11 terms of the arithmetic progression is 88, so: \[ S_{11} = \frac{11}{2} \left( 2a + 10d \right) = 88. \] Substitute \( d = \frac{3}{2} \): \[ \frac{11}{2} \left( 2a + 15 \right) = 88 \quad \Rightarrow \quad 11(2a + 15) = 176 \quad \Rightarrow \quad 2a + 15 = 16 \quad \Rightarrow \quad 2a = 1 \quad \Rightarrow \quad a = \frac{1}{2}. \] Now, the roots of the quadratic equation \( 3x^2 - px + q = 0 \) are the 10th and 11th terms of the arithmetic progression. Therefore, the sum of the roots is \( \frac{p}{3} \), and the product of the roots is \( \frac{q}{3} \). The sum of the roots is: \[ T_{10} + T_{11} = (a + 9d) + (a + 10d) = 2a + 19d = 2 \cdot \frac{1}{2} + 19 \cdot \frac{3}{2} = 1 + 28.5 = 29.5. \] Thus: \[ \frac{p}{3} = 29.5 \quad \Rightarrow \quad p = 88.5. \] The product of the roots is: \[ T_{10} \cdot T_{11} = (a + 9d)(a + 10d) = a^2 + 19ad + 90d^2 = \left( \frac{1}{2} \right)^2 + 19 \cdot \frac{1}{2} \cdot \frac{3}{2} + 90 \cdot \left( \frac{3}{2} \right)^2 = \frac{1}{4} + \frac{57}{4} + \frac{405}{4} = \frac{463}{4}. \] Thus: \[ \frac{q}{3} = \frac{463}{4} \quad \Rightarrow \quad q = \frac{463}{4} \times 3 = 348.25. \] Finally, we calculate \( q - 2q \): \[ q - 2q = 348.25 - 696.5 = 474. \]