Question

For an a.p 1/6,1/4,1/3... Find t20 and s10​

Answer 1

Answer:

1/6 , 1/4 , 1/3 ,... are in A.P

first term = a = a1 = 1/6

common difference ( d ) = a2 - a1

d = 1/4 - 1/6

d = ( 3 - 2 ) / 12

d = 1/2

we know that

1 ) n th term in A.P = tn = a + ( n - 1 )d

2 ) sum of n terms = Sn = n/2[ 2a + ( n - 1 )d ]

1 ) a = 1/6 , d = 1/2 , n = 20

t20 = a + 19d

= 1/6 + 19 ( 1/2)

= 1/6 + 19/2

= ( 1 + 57 ) / 6

= 58/6

= 29/3

2 ) a = 1/6 , d = 1/2 , n = 10

S10 = ( 10/2 ) [ 2 × 1/6 + 9 × 1/2 ]

= 5 [ 1/3 + 9/2 ]

= ( 5/6 ) ( 2 + 27 )

= ( 5 × 29 ) / 6

= 145/6