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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We have been given that the 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 13thterm of an AP.

Sum of the first n terms is given by

Sn=2n[2a+(n−1)d]

S6=26[2a+(6−1)d]

S6=3[2a+5d]

T30T10=31

a+29da+9d=31

(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+(n−1)d

=2+(13−1)2

=2+24

=26