Question

find next four terms of sequence 1/6,1/4,1/3 also find Sn​

Answer 1

Given  : 1/6,1/4,1/3  

To Find : next four terms of sequence

Sn​

Solution:

1/4 - 1/6  =  ( 3 - 2)/12 = 1/12

1/3 - 1/4  = (4 - 3)/12 = 1/12

d = 1/12

1/6 = 2/12

1/4 = 3/12

1/3  = 4/12

Next 4 terms will be

5/12

6/12 = 1/2

7/12

8/12  = 2/3

next four terms of sequence  = 5/12 , 1/2 , 7/12 , 2/3

Sum on n terms

=2/12 + 3/12  + 4/12 +  +  +  +  (n+1)/12

= (1/12) (2 + 3 + 4 +  +  + +  (n + 1))

= (1/12) ( n/2)(2 + n + 1)

= n(n + 3)/24

n(n + 3)/24  is Sn

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Answer 2

To Find :-

• Next four terms of sequence 1/6,1/4,1/3 .
• Sn = ?

Solution :-

we have,

• First term = a = 1/6 .
• Common difference = d = a2 - a1 = 1/4 - 1/6 = (3-2)/12 = 1/12 .

we know that,

• An = a + (n - 1)d .
• Sn = (n/2)[2a + (n - 1)d] .

so,

→ A(4) = a + (4 - 1)d

→ A(4) = 1/6 + 3 * (1/12)

→ A(4) = 1/6 + 1/4

→ A(4) = (2 + 3)/12 = 5/12 .

similarly,

→ A(5) = a + (5 - 1)d

→ A(5) = 1/6 + 4 * (1/12)

→ A(5) = 1/6 + 4/12

→ A(5) = (2 + 4)/12 = 6/12 = 1/2 .

and,

→ A(6) = a + (6 - 1)d

→ A(6) = 1/6 + 5 * (1/12)

→ A(6) = 1/6 + 5/12

→ A(6) = (2 + 5)/12 = 7/12 .

also,

→ A(7) = a + (7 - 1)d

→ A(7) = 1/6 + 6 * (1/12)

→ A(7) = 1/6 + 6/12

→ A(7) = (2 + 6)/12 = 8/12 = 2/3 .

therefore,

→ Sn = (n/2)[2a + (n - 1)d]

→ Sn = (n/2)[2*1/6 + (n - 1)(1/12)]

→ Sn = (n/2)[1/3 + n/12 - 1/12]

→ Sn = (n/2)(4 + n - 1)/12

→ Sn = (n/2)(3 + n)/12

→ Sn = n(3 + n)/24 (Ans.)