The roots of the quadratic equation 3 x²-p x+q=0 are 10ᵗʰ and 11ᵗʰ terms of an arithmetic progression with common difference 3/2. If the sum of the first 11 terms of this arithmetic progression is 88 , then q-2 p is equal to ________ .

Solution

Maths sir · 4 steps

The sum of 11 terms is 11 times the middle term. That gives T₆, and T₁₀ and T₁₁ follow from it.

1. Middle term. For 11 terms, S₁₁ = 11/2(2a + 10d) = 11(a + 5d) = 11 T₆. So 11 T₆ = 88, which gives T₆ = 8.
2. Find the roots. With d = 3/2, T₁₀ = T₆ + 4d = 8 + 6 = 14 and T₁₁ = T₁₀ + d = 14 + 1.5 = 15.5. These are the two roots.

   α + β = p/3, αβ = q/3

3. Use the coefficients. The sum of the roots is 14 + 15.5 = 29.5 = p/3, so p = 88.5. The product is 14 × 15.5 = 217 = q/3, so q = 651.
4. So. q - 2p = 651 - 177 = 474.

The common slip: 158. Taking the sum of the roots as p and the product as q, forgetting the leading coefficient 3, gives p = 29.5, q = 217 and q - 2p = 158, not 474.

The idea to keep. An AP with an odd number of terms has its sum equal to the number of terms times the middle term. And always divide by the leading coefficient before reading off the sum and product of the roots.

Answer from the NTA final answer key. Solution by PrepWiser.