Question

the sum of first 6 term of an ap is 42 the ratio of 10th term to 30 term is 1 ratio 3 calculate the first and 13th term of AP

Answer 1

S6 =42

a + 9d                    1
------------     =       -------
a + 29d                  3

cross multiply we  get

3a + 27d = a +29 d

2a - 2d = 0 ------------------ (1)
 
its given that 
sum of first six terms of an AP is 42

therefore 

S6 = n/2 ( 2a + (n-1) d)

42 = 6/2 ( 2a + (6-1) d)

42 = 3 (2a + 5d )

14 = 2a +5d

2a +5d = 14 ------------------- (2)

solve eq 1 and 2

2a - 2d = 0
2a +5d = 14

we get 
d= 2
a = 2
---------------
13th term of AP
= a + (n-1)d

2+ (13-1) 2

= 2+ 24

AP=26

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