Exams Joint Entrance Examination Mathematics Sequences and Series Question 95 Question 95 of 519 Medium NUMERICAL
The roots of the quadratic equation 3 x 2 − p x + q = 0 3{x}^{2}-px+q=0 3 x 2 − p x + q = 0 are 10 {10}^{} 10 and 11 {11}^{} 11 terms of an arithmetic progression with common difference 3 2 \frac{3}{2} 2 3 . If the sum of the first 11 terms of this arithmetic progression is 88 , then q − 2 p q-2p q − 2 p is equal to \_\_\_\_\_\_\_\_ .
Answer: 474
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S 11 = 11 2 ( 2 a + 10 d ) = 88 a + 5 d = 8 a = 8 − 5 × 3 2 = 1 2 T 10 = a + 9 d = 1 2 + 9 × 3 2 = 14 T 11 = a + 10 d = 1 2 + 10 × 3 2 = 31 2 p 3 = T 10 + T 11 = 14 + 31 2 = 59 2 p = 177 2 q 3 = T 10 × T 11 = 7 × 31 = 217 q = 651 q − 2 p = 651 − 177 = 474 \begin{gathered} {S}_{11}=\frac{11}{2}(2a+10d)=88 \\ a+5d=8 \\ a=8-5\times \frac{3}{2}=\frac{1}{2} \\ \\ {T}_{10}=a+9d=\frac{1}{2}+9\times \frac{3}{2}=14 \\ {T}_{11}=a+10d=\frac{1}{2}+10\times \frac{3}{2}=\frac{31}{2} \\ \frac{p}{3}={T}_{10}+{T}_{11}=14+\frac{31}{2}=\frac{59}{2} \\ p=\frac{177}{2} \\ \frac{q}{3}={T}_{10}\times {T}_{11}=7\times 31=217 \\ q=651 \\ q-2p \\ =651-177 \\ =474 \end{gathered} S 11 = 2 11 ( 2 a + 10 d ) = 88 a + 5 d = 8 a = 8 − 5 × 2 3 = 2 1 T 10 = a + 9 d = 2 1 + 9 × 2 3 = 14 T 11 = a + 10 d = 2 1 + 10 × 2 3 = 2 31 3 p = T 10 + T 11 = 14 + 2 31 = 2 59 p = 2 177 3 q = T 10 × T 11 = 7 × 31 = 217 q = 651 q − 2 p = 651 − 177 = 474 Answer the question to unlock