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djyliett [7]
4 years ago
13

What is the sum to 'n' terms of the series √5,√20,√45,√80,.....?

Mathematics
1 answer:
Yakvenalex [24]4 years ago
7 0

Answer:

Sₙ = √5/2 n (n + 1)

Step-by-step explanation:

<u>Given AP:</u>

  • √5,√20,√45,√80,...

<u>Rewriting as:</u>

  • √5,√4*5,√9*5,√16*5,...
  • √5, 2√5, 3√5, 4√5,..., n√5

<u>Common difference:</u>

  • d= 2√5 - √5 = √5

<u>Sum of n terms:</u>

  • Sₙ = 1/2n (a₁ + aₙ) =
  • 1/2n (√5 + √5 + (n-1)√5) =
  • 1/2n (n + 1)√5=
  • √5/2 n (n + 1)

Sₙ = √5/2 n (n + 1)

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9 cos(sin¯¹(x)) = √81 – 81x²
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The equation 9cos(sin¯¹(x)) = √(81 – 81x²) is true since L.H.S = R.H.S

To answer the question, we need to know what an equation is

<h3>What is an equation?</h3>

An equation is a mathematical expression that show the relationship between two variables.

Given 9cos(sin¯¹(x)) = √(81 – 81x²), we need to show L.H.S = R.H.S

So, L.H.S = 9cos(sin¯¹(x))

= 9[√{1 - sin²(sin¯¹(x)}] (Since sin²y + cos²y = 1 ⇒ cosy = √[1 - sin²y])

9[√{1 - sin²(sin¯¹(x)}] = √9² × √{1 - sin²(sin¯¹(x)}]

= √[9² × {1 - sin²(sin¯¹(x)}]

= √[81 × {1 - sin²(sin¯¹(x)}]

= √[81 × {1 - x²}]   (since sin²(sin¯¹(x) = [sin(sin¯¹(x)]² = x²)

= √(81 – 81x²)

= R.H.S

So, the equation 9cos(sin¯¹(x)) = √(81 – 81x²) is true since L.H.S = R.H.S

Learn more about equations here:

brainly.com/question/2888445

#SPJ1

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