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Question

Sum of first six terms of an AP is 42. The ratio of its 10th term to 30th term is 1:3 find the first and 13th term of an AP.

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Solution

Sum of the first n terms is given by

Sn=n2[2a+(n1)d]

S6=62[2a+(61)d]

S6=3[2a+5d]

T10T30=13

a+9da+29d=13

(a+9d)3=(a+29d)1

3a+27d=a+29d

2a=2d

a=d...(1)

S6=3[2a+5d]

S6=3[2a+5a]

42=3(7a)

42=21a

a=2

From (1)

d=a=2

T13=a+(n1)d

=2+(131)2

=2+24

=26

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