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Let [tex]\sf a+b,a-b,ab,\dfrac{a}{b}[/tex] form an arithmetic sequence. What is the sixth term of this sequence?​

Respuesta :

mhanifa

Answer:

  • 171/40 or 4 11/40

Step-by-step explanation:

AP given

  • a + b, a - b,  ab, a/b

To find

  • 6th term

Solution

Common difference

Difference of first two

  • d = (a -b) - (a + b) = -2b

Difference of second two

  • d= ab - (a - b)

Difference of last two

  • d = a/b - ab

Now comparing d:

  • -2b = ab - (a - b)
  • ab - a = - 3b
  • a(1 - b) = 3b
  • a = 3b/(1 - b)

and

  • a/b - ab = -2b
  • a(1/b - b) = -2b
  • a = 2b²/(b² - 1)

Eliminating a:

  • 2b²/(b² - 1) = 3b/(1 - b)
  • 2b/(b+1) = -3
  • 2b = -3b - 3
  • 5b = - 3
  • b = -3/5

Finding a:

  • a = 3b/(1 - b) =
  • 3*(-3/5) *1/(1 - (-3/5)) =
  • -9/5*5/8 =
  • -9/8

So the first term is:

  • a + b = -3/5 - 9/8 = -24/40 - 45/40 = - 69/40

Common difference:

  • d = -2b = -2(-3/5) = 6/5

The 6th term:

  • a₆ = a₁ + 5d =
  • -69/40 + 5*6/5 =
  • -69/40 + 240/40 =
  • 171/40 = 4 11/40

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