SATHEE: Sequences and Series 2 Question 14

Sequences and Series 2 Question 14

15.

Let $p$ and $q$ be the roots of the equation $x^{2} - 2 x + A = 0$ and let $r$ and $s$ be the roots of the equation $x^{2} - 18 x + B = 0$ . If $p < q < r < s$ are in arithmetic progression, then $A = \dots$ and $B = \dots$ .

(1997, 2M)

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Answer:

Correct Answer: 15. $(A = - 3, B = 77)$

Solution:

Given, $p + q = 2, p q = A$

and $r + s = 18, r s = B$

and it is given that $p, q, r, s$ are in an AP.

Therefore, let $p = a - 3 d, q = a - d, r = a + d$

and $s = a + 3 d$

Since, $p < q < r < s$

We have, $d > 0$

$\begin{aligned} Now, 2 & = p + q = a - 3 d + a - d = 2 a - 4 d \Rightarrow a - 2 d & = 1 \end{aligned}$

Again, $18 = r + s = a + d + a + 3 d$

$18 = 2 a + 4 d$

$\Rightarrow 9 = a + 2 d$

On subtracting Eq. (i) from Eq. (ii), we get

$8 = 4 d \Rightarrow d = 2$

On putting in Eq. (ii), we get $a = 5$

$\begin{aligned} ∴ p & = a - 3 d = 5 - 6 = - 1 \\ q & = a - d = 5 - 2 = 3 \\ r & = a + d = 5 + 2 = 7 \end{aligned}$

and $s = a + 3 d = 5 + 6 = 11$

Therefore, $A = p q = - 3$ and $B = r s = 77$

Sequences and Series 2 Question 14

15.

Answer:

Engineering Preparation

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Sequences and Series 2 Question 14

15.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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