If x2-x-1-x2-2x-3-x2-3x-5-x2-20x-39-4500- then x is equal to

The correct options are A 10 D 20.5(x^2+x+1)+(x^2+2x+3)+(x^2+3x+5)+⋯+(x^2+20x+39)=4500 ⇒ 20x^2+x(1+2+3+⋯+20)+(1+3+⋯+39)=4500 ⇒ 20x^2+x (20·212)+20^2=4500 ⇒ x^2 ...

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Byju's Answer

Standard XII

Mathematics

Sigma n

If x2+x+1+x2+...

Question

If $(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots + (x^{2} + 20 x + 39) = 4500,$ then $x$ is equal to

10

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- 10

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20.5

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- 20.5

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Solution

The correct options are
A $10$
D $- 20.5$
$(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots + (x^{2} + 20 x + 39) = 4500$
$\Rightarrow 20 x^{2} + x (1 + 2 + 3 + \dots + 20) + (1 + 3 + \dots + 39) = 4500$
$\Rightarrow 20 x^{2} + x (\frac{20 \cdot 21}{2}) + 20^{2} = 4500$
$\Rightarrow x^{2} + \frac{21}{2} x + 20 = 225$
$\Rightarrow 2 x^{2} + 21 x - 450 + 40 = 0$
$\Rightarrow 2 x^{2} + 21 x - 410 = 0$
$\Rightarrow (x - 10) (2 x + 41) = 0$
$∴ x = 10$ or $x = - 20.5$

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Similar questions

Q. Solve the following equation:

(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots \dots + (x^{2} + 20 x + 39) = 4500.

Q. If

(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots + (x^{2} + 20 x + 39) = 4500,

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Solve the following equation:

\frac{2 x - 3}{x^{2} - 2 x + 3} = \frac{3 x - 5}{x^{2} - 3 x + 5}

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If (x2+x+1)+(x2+2x+3)+(x2+3x+5)+⋯+(x2+20x+39)=4500, then x is equal to

The correct options are A 10 D −20.5(x2+x+1)+(x2+2x+3)+(x2+3x+5)+⋯+(x2+20x+39)=4500 ⇒20x2+x(1+2+3+⋯+20)+(1+3+⋯+39)=4500 ⇒20x2+x(20⋅212)+202=4500 ⇒x2+212x+20=225 ⇒2x2+21x−450+40=0 ⇒2x2+21x−410=0 ⇒(x−10)(2x+41)=0 ∴x=10 or x=−20.5

If $(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots + (x^{2} + 20 x + 39) = 4500,$ then $x$ is equal to