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dolphi86 [110]
3 years ago
8

The fourth term of an arithmetic

Mathematics
1 answer:
yulyashka [42]3 years ago
5 0

Answer:

2

Step-by-step explanation:

<u>Given AP where:</u>

  • a₄ = 2a₂ - 1
  • a₆ = 7

<u>To find</u>

  • a₁ = ?

<u>Since</u>

  • a₄ = a₁ + 3d
  • a₂ = a₁ + d
  • a₆ = a₁ + 5d

<u>Initial equations will change as:</u>

  • a₁ + 3d = 2(a₁ + d) - 1  ⇒  a₁ + 3d = 2a₁ + 2d - 1 ⇒ a₁ = d + 1
  • a₁ + 5d = 7 ⇒ a₁ = 7 - 5d

<u>Comparing the above:</u>

  • d + 1 = 7 - 5d
  • 6d = 6
  • d = 1

<u>Then:</u>

  • a₁ = d + 1 = 1 + 1 = 2
  • a₁ = 2

The first term is 2

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If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.

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