Exams Joint Entrance Examination Mathematics Sequences and Series Question 142 Question 142 of 519 Medium NUMERICAL
The roots of the quadratic equation 3 x 2 − p x + q = 0 3x^2 - px + q = 0 3 x 2 − p x + q = 0 are the 10 10 10 th and 11 11 11 th terms of an arithmetic progression with common difference 3 2 \dfrac32 2 3 . If the sum of the first 11 11 11 terms of this arithmetic progression is 88 88 88 , then q − 2 p q - 2p q − 2 p is equal to \_\_\_\_\_\_\_\_.
Answer: 474
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\[S_{11} = \frac{11}{2}(2a+10d) = 88 \implies a+5d = 8 \implies a = 8 - 5\left(\frac32\right) = \frac12.\]
\[T_{10} = a+9d = \frac12+\frac{27}{2} = 14, \qquad T_{11} = a+10d = \frac12+15 = \frac{31}{2}.\]
By Vieta's formulas for 3 x 2 − p x + q = 0 3x^2-px+q=0 3 x 2 − p x + q = 0 : p 3 = T 10 + T 11 = 14 + 31 2 = 59 2 ⇒ p = 177 2 \dfrac{p}{3}=T_{10}+T_{11}=14+\dfrac{31}{2}=\dfrac{59}{2} \Rightarrow p=\dfrac{177}{2} 3 p = T 10 + T 11 = 14 + 2 31 = 2 59 ⇒ p = 2 177 , and q 3 = T 10 ⋅ T 11 = 14 × 31 2 = 217 ⇒ q = 651 \dfrac{q}{3}=T_{10}\cdot T_{11}=14\times\dfrac{31}{2}=217 \Rightarrow q=651 3 q = T 10 ⋅ T 11 = 14 × 2 31 = 217 ⇒ q = 651 .
q − 2 p = 651 − 177 = 474. q-2p = 651 - 177 = 474. q − 2 p = 651 − 177 = 474. Answer the question to unlock