SATHEE: Theory of Equations 1 Question 14

Theory of Equations 1 Question 14

15. For a positive integer $n$ , if the quadratic equation, $x (x + 1) + (x + 1) (x + 2) + \dots + (x + \overset{―}{n - 1}) (x + n) = 10 n$ has two consecutive integral solutions, then $n$ is equal to

(2017 Main)

(a) 12

(b) 9

(d) 11

Show Answer

Answer:

Correct Answer: 15. (d)

Solution:

Given quadratic equation is

$\begin{array}{cc} x (x + 1) + (x + 1) (x + 2) + \dots + (x + \overset{―}{n - 1}) (x + n) = 10 n \\ \Rightarrow & (x^{2} + x^{2} + \dots + x^{2}) + [(1 + 3 + 5 + \dots + (2 n - 1)] x \\ + [(1 \cdot 2 + 2 \cdot 3 + \dots + (n - 1) n] = 10 n \\ \Rightarrow & n x^{2} + n^{2} x + \frac{n (n^{2} - 1)}{3} - 10 n = 0 \\ \Rightarrow & x^{2} + n x + \frac{n^{2} - 1}{3} - 10 = 0 \\ \Rightarrow & 3 x^{2} + 3 n x + n^{2} - 31 = 0 \end{array}$

Let $α$ and $β$ be the roots.

Since, $α$ and $β$ are consecutive.

$∴ | α - β | = 1 \Rightarrow (α - β)^{2} = 1$

Again, $(α - β)^{2} = (α + β)^{2} - 4 α β$

$\begin{array}{lcl} \Rightarrow & 1 = {\frac{- 3 n}{3}}^{2} - 4 \frac{n^{2} - 31}{3} \\ \Rightarrow & 1 = n^{2} - \frac{4}{3} (n^{2} - 31) \Rightarrow 3 = 3 n^{2} - 4 n^{2} + 124 \\ \Rightarrow & n^{2} = 121 \Rightarrow n = \pm 11 \\ ∴ & n = 11 \end{array}$

$[∵ n > 0]$

Theory of Equations 1 Question 14

15. For a positive integer $n$ , if the quadratic equation, $x (x + 1) + (x + 1) (x + 2) + \dots + (x + \overset{―}{n - 1}) (x + n) = 10 n$ has two consecutive integral solutions, then $n$ is equal to

Answer:

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Theory of Equations 1 Question 14

15. For a positive integer n, if the quadratic equation, x(x+1)+(x+1)(x+2)+…+(x+n−1―)(x+n)=10n has two consecutive integral solutions, then n is equal to

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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15. For a positive integer $n$ , if the quadratic equation, $x (x + 1) + (x + 1) (x + 2) + \dots + (x + \overset{―}{n - 1}) (x + n) = 10 n$ has two consecutive integral solutions, then $n$ is equal to