Question

Let
`\sf a+b,a-b,ab,(a)/(b)` form an arithmetic sequence. What is the sixth term of this sequence?​

Answer 1

• 171/40 or 4 11/40

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