The sum of first six terms of an AP is 42.The ratio of its 10th term to its 30th term is 1:3 .Calculate the first term and thirteenth term of an AP
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Answer:
Sum of the first n terms is given by
S
n
=
2
n
[2a+(n−1)d]
S
6
=
2
6
[2a+(6−1)d]
S
6
=3[2a+5d]
T
30
T
10
=
3
1
a+29d
a+9d
=
3
1
(a+9d)3=(a+29d)1
3a+27d=a+29d
2a=2d
a=d...(1)
S
6
=3[2a+5d]
S
6
=3[2a+5a]
42=3(7a)
42=21a
∴a=2
From (1)
d=a=2
T
13
=a+(n−1)d
=2+(13−1)2
=2+24
=26
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