Question

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

Answer 1

S6 = 42
6/2 {2a + (6-1) d} = 42
3 {2a + 5d} = 42
{2a + 5d} = 42/3
{2a + 5d} = 14..........(1)

T10 = a +9d = 1 divide by
T30 = a + 29d = 3
3(a + 9d) = 1(a+29d)
3a + 27d = a + 29d
3a-a + 27d - 29d = 0
2a - 2d = 0
a - d = 0/2
a = d..............(2)
Putting (2) equ. in (1) equ.
2(d) + 5d = 14
7d = 14
d = 2
Therefore a=d=2

Thus, T13 = a + 12d
= 2 + 12(2)
= 26

SO, the 13th term is 26.