Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 57521 by rahul 19 last updated on 06/Apr/19

Commented by rahul 19 last updated on 06/Apr/19

$A n s - (a) 11 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Apr/19
	$T_n =x^2 +nx+(n−1)x+(n^2 −n) T_n =x^2 +(2n−1)x+(n^2 −n) T_1 =x^2 +(2.1−1)x+(1^2 −1) T_2 =x^2 +(2.2−1)x+(2^2 −2) T_3 =x^2 +(2.3−1)x+(3^2 −3) ... ... on addition ΣT_n =nx^2 +[2.(n/2){2.1+(n−1)1}−n]x+((n(n+1)(2n+1))/6)−((n(n+1))/2) =nx^2 +[n(n+1)−n]x+((n(n+1))/2)(((2n+1)/3)−1) =nx^2 +(n^2 x)+((n(n+1))/2)(((2n−2)/3)) =nx^2 +n^2 x+(n/3)(n^2 −1) so the eqn is nx^2 +n^2 x+(n/3)(n^2 −1)=10n x^2 +nx+((n^2 −1)/3)−10=0 given β=α+1 (α−β)^2 =(α+β)^2 −4αβ 1=(−n)^2 −4[((n^2 −1)/3)−10] 1=n^2 −4(((n^2 −1−30)/3)) 3=3n^2 −4n^2 +124 n^2 =121 n=11$
	$T_{n} = x^{2} + n x + (n - 1) x + (n^{2} - n)$ $T_{n} = x^{2} + (2 n - 1) x + (n^{2} - n)$ $T_{1} = x^{2} + (2.1 - 1) x + (1^{2} - 1)$ $T_{2} = x^{2} + (2.2 - 1) x + (2^{2} - 2)$ $T_{3} = x^{2} + (2.3 - 1) x + (3^{2} - 3)$ $. . .$ $. . .$ $o n a d d i t i o n$ $Σ T_{n} = {n x}^{2} + [2 . \frac{n}{2} {2.1 + (n - 1) 1} - n] x + \frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2}$ $= {n x}^{2} + [n (n + 1) - n] x + \frac{n (n + 1)}{2} (\frac{2 n + 1}{3} - 1)$ $= {n x}^{2} + (n^{2} x) + \frac{n (n + 1)}{2} (\frac{2 n - 2}{3})$ $= {n x}^{2} + n^{2} x + \frac{n}{3} (n^{2} - 1)$ $s o t h e e q n i s$ ${n x}^{2} + n^{2} x + \frac{n}{3} (n^{2} - 1) = 10 n$ $x^{2} + n x + \frac{n^{2} - 1}{3} - 10 = 0$ $g i v e n β = α + 1$ ${(α - β)}^{2} = {(α + β)}^{2} - 4 α β$ $1 = {(- n)}^{2} - 4 [\frac{n^{2} - 1}{3} - 10]$ $1 = n^{2} - 4 (\frac{n^{2} - 1 - 30}{3})$ $3 = 3 n^{2} - 4 n^{2} + 124$ $n^{2} = 121$ $n = 11$
	Commented by rahul 19 last updated on 07/Apr/19

	$t h a n k s s i r!$
	Commented by peter frank last updated on 07/Apr/19

	$t h a n k y o u$
	Answered by Rasheed.Sindhi last updated on 07/Apr/19

	$x (x + 1) + (x + 1) (x + 2) + . . . (x + \overset{―}{n - 1}) (x + n) = 10 n$ ${n x}^{2} + n^{2} x + \frac{n}{3} (n^{2} - 1) = 10 n$ $[F r o m t a n m a y s i r]$ $L e t α & (α + 1) a r e s o l u t i o n s :$ $α (α + 1) + (α + 1) (α + 2) + . . . (α + \overset{―}{n - 1}) (α + n) = 10 n . . . (I)$ $(α + 1) (α + 2) + (α + 2) (α + 3) . . . (α + n) (α + \overset{―}{n + 1}) = 10 n . . . (I I)$ $(I) - (I I) :$ $α (α + 1) - (α + n) (α + n + 1) = 0$ $α^{2} + α - [α^{2} + (2 n + 1) α + n (n + 1)] = 0$ $- 2 n α - n^{2} - n = 0$ $n^{2} + (2 α + 1) n = 0$ $n (n + 2 α + 1) = 0$ $A s n > 0 \Rightarrow n + 2 α + 1 = 0$ $n = - 2 α - 1$ $A c c o r d i n g t h e g u i d a n c e o f s i r m r W :$ $α = - \frac{n + 1}{2}$ $P u t t i n g i n (I)$ $(- \frac{n + 1}{2}) (- \frac{n + 1}{2} + 1) + (- \frac{n + 1}{2} + 1) (- \frac{n + 1}{2} + 2) + . . . (- \frac{n + 1}{2} + \overset{―}{n - 1}) (- \frac{n + 1}{2} + n) = 10 n . . . (I)$ $O r e q u i v a l e n t l y a c c o r d i n g t o s i r t a n m a y$ $n {(- \frac{n + 1}{2})}^{2} + n^{2} (- \frac{n + 1}{2}) + \frac{n}{3} (n^{2} - 1) = 10 n$ $\frac{n {(n + 1)}^{2}}{4} - \frac{n^{2} (n + 1)}{2} + \frac{n (n^{2} - 1)}{3} = 10 n$ $3 n {(n + 1)}^{2} - 6 n^{2} (n + 1) + 4 n (n^{2} - 1) = 120 n$ $3 n^{3} + 6 n^{2} + 3 n - 6 n^{3} - 6 n^{2} + 4 n^{3} - 4 n - 120 n = 0$ $n^{3} - 121 n = 0$ $n (n^{2} - 121) = 0$ $∵ n > 0 \Rightarrow n^{2} = 121 \Rightarrow n = 11$
	Commented by mr W last updated on 07/Apr/19

	$y o u a r e n o t w r o n g s i r . y o u h a v e t w o$ $u n k n o w n s α a n d n, a n d y o u h a v e t w o$ $e q u a t i o n s I a n d I I . n o w y o u h a v e g o t$ $n = - 2 α - 1 o r α = - \frac{n + 1}{2}, y o u c a n p u t$ $t h i s i n t o I t o f i n d n .$
	Commented by Rasheed.Sindhi last updated on 07/Apr/19

	$θ N X S i R m r W!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum