Question

term of the sequence 1/6, 1/4, 1/3 and also find Sn

Answer 1

Answer:

Step-by-step explanation:

Let a = 1/6, a₂ = 1/4, a₃ = 1/3

a₂ - a = 1/4 - 1/6 = 1/12

a₃ - a₂ = 1/3 - 1/4 = 1/12

Thus,  this is an AP with a common difference 'd' = 1/12

We know, aₙ = a + (n-1)d

So, the next four terms are

a₄ = a + 3d => 1/6 + 3*1/12 => 1/6 + 1/4 = 5/12

a₅ = a + 4d => 1/6 + 4* 1/12 => 1/6 + 1/3 = 1/2

a₆ = a + 5d => 1/6 + 5 * 1/12=> 1/6 + 5/12 =7/12

a₇ = a + 6d => 1/6 + 6 * 1/12 => 1/6 + 1/2 = 2/3

Sₙ = n(2a + (n-1)d)/2

Sₙ = n(1/3 + n/12 - 1/12)/2

Sₙ = n(1/4 + n/12)/2

Sₙ = n/4(1 + n/3)/2

Sₙ = (n + n²/3)/8

Sₙ = 3n + n²/24