Question

20

Let a₁,a₂,a₃⋯,a₂₁ be an AP such that Σ 1/aₙaₙ+1= 4/9 If the sum of this AP is 189, then a₆a₁₆ is equal to:
n=1

A. 36
B. 57
C. 72
D. 48

Answer 1

To find a₆a₁₆, we first find the common difference of the AP. Then, we use the formulas for the nth term and sum of the AP to find a₁ and a₂₀. Substituting these values, we calculate a₆ and a₁₆. Finally, we find a₆a₁₆ by multiplying the two values.

Step-by-step explanation:

To find the value of a₆a₁₆, we first need to find the common difference (d) of the arithmetic progression (AP).

We are given that Σ 1/aₙaₙ+1 = 4/9. Using the formula for the sum of the reciprocal products of consecutive terms in an AP, we get:

Σ 1/aₙaₙ+1 = (a₁/a₂)(a₂/a₃)(a₃/a₄)…(a₂₀/a₂₁) = 4/9

Since the sum of the AP is 189, we know that (a₁ + a₂₀)/2 * 20 = 189.

Solving these two equations simultaneously, we find that a₁ = 15 and a₂₀ = 255.

Now, we can calculate a₆a₁₆ by using the formula for the nth term of an AP (aₙ = a₁ + (n-1)d):

a₆ = a₁ + 5d and a₁₆ = a₁ + 15d.

Substituting the values of a₁ and a₂₀ into these equations, we get:

a₆ = 15 + 5d and a₁₆ = 15 + 15d

Since the sum of the AP is 189, we have the equation:

189 = (a₁ + a₂₀)/2 * 20 = (15 + 255)/2 * 20

Solving for d, we find d = 6.

Substituting this value back into the equations for a₆ and a₁₆, we get:

a₆ = 15 + 5(6) = 45 and a₁₆ = 15 + 15(6) = 105.

Finally, we calculate a₆a₁₆: a₆a₁₆ = 45 * 105 = 4725.