Question

For an A.P. 1/6,1/4,1/3,...... Find t20 & S10

Answer 1

Hi ,

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

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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 ]

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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

I hope this helps you.

:)

Answer 2

A.P,

1/6,1/4,1/3.....

here, a=1/6

d=t2-t1. or d=t3-t4

= 1/4-1/6. or = 1/3-1/4

= 2/24. or = 3-2/12

d = 1/12. or d = 1/12

tn= a+(n-1)d

t20=1/6+(20-1)1/12

=1/6+(19)1/12

=1/6+19/12

=2/12+19/12....... multiply 1/6 by 2/2

=2+19/12

=21/12

t20= 5.415

Sn=n/2(2a+(n-1)d)

=10/2(2×1/6+(20-1)1/12)

=5(1/3+(19)1/12)

=5(1/3+19/12)

=5(4/12+19/12)....multiply 1/3 by4/4

=5(4+19/12)

=5(23/12)

=5(1.915)

Sn=9.575