Let a1,a2 ,......,a21 be an AP such that ∑1/(anan+1),n∈​​​​​​​(1,20) = 4/9. If the sum of this AP is 189, then a6a16 is equal to :

Question

Let a₁,a₂ ,......,a₂₁ be an AP such that \(\displaystyle\sum_{n=1}^{20}\frac{1}{a_na_{n+1}}\) = \(\frac{4}{9}\). If the sum of this AP is 189, then a₆a₁₆ is equal to :

∑1/(a_na_n+1),n∈(1,20) = 4/9

(1) 57

(2) 72

(3) 48

(4) 36

Answer 1

Let first term be a and common difference d

\(\frac 1{a_1a_2} + \frac 1{a_2a_3} + .....+ \frac 1{a_{20}a_{21}} = \frac 49\)  ......(i)

Also, a₁ + a₂ + ⋯ + a₂₁ = 189   ......(ii)

By (i),

\(\frac 1{a_1} - \frac 1{a_2} + \frac 1{a_2} - \frac 1{a_3} + ....+ \frac 1{a_{20}} - \frac 1{a_{21}} = \frac {4d}{9}\)

⇒ \(\frac 1a - \frac 1{a + 20d} = \frac{4d}9\)

⇒ \(\frac{20d}{a(a + 20d)} = \frac {4d}9\)

⇒ 45 = a(a + 20d)  .......(iii)

From (ii),

21a + 210d = 189

⇒ a + 10d = 9    ......(iv)

By (iii) and (iv),

d = \(\frac 35\) and a = 3

∴ a₆a₁₆ = (3 + 3)(3 + 9) = 72

Answer 2

Option : (2). 72

∑1/(a_na_n+1),n∈(1,20)

\(\displaystyle\sum_{n=1}^{20}\frac{1}{a_na_{n+1}}\)= \(\displaystyle\sum_{n=1}^{20}\frac{1}{a_n(a_n+d)}\) 

Now,

Sum of first 21 terms = \(\frac{21}{2}\)(2a₁ + 20d) = 189

⇒ a₁ + 10d = 9 ...(2)

For equation (1) & (2) we get,

a₁ = 3 & d = \(\frac{3}{5}\)

OR

a₁ = 15 & d = \(-\frac{3}{5}\)

So, 

a₆ .a₁₆ = (a₁ + 5d) (a₁ + 15d)

⇒ a₁a₁₆ = 72