Question

find t20 and S10 from 1/6 , 1/4, 1/3​

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 the 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 + 1/2 ]

= ( 5/6) ( 2+27)

= ( 5× 29) / 6

= 145/6