Sum of first six terms of an AP is 42 and ratio of 10th term to its 30th term is 1:3. Calculate fist and 13th term?

Question

An arithmetic sequence formed of 11 terms, and the sum of all its terms equals to 220. Find the middle term in that sequence.

Answer 1

Tn = [a + (n-1)d]
Sn = n/2*[2a +(n-1)d].
Given
T30/T10 = 3
(a + 29d)/(a + 9d) = 3
After simplification we get
a = d

S6 = 6/2 [2a + 5d] = 42 (given)
14 = 2a + 5d
14/7 = 2 = a = d ( substituting a = d)

So the series starts like
2, 4, 6, 8, 10, 12, 14, 16, .......
Now T13 = [2+12d] = [2 + 24] = 26.