Question

The sum of first 6 term of an ap is 42.The ratio of 10th term to 30th term is 1/3.Calculate it's 13th term.

Answer 1

26 is the answer.
Explanation:

Let a1 = the first term
Then we have, the sum of the 1st 6 terms = 42.

Therefore,

42 = (6/2)[2a1 + 5d]

42 = 3[2a1 + 5d]

14 = 2a1 + 5d → a1 = [14 - 5d]/2

And a30 = 3(a10)
Therefore,

3(a10) = a10 + 20d

2(a10) = 20d

a10 = 10d

So we have,

a10 = (a1 + 9d)
And by substitution,

10d = ([14- 5d]/2 + 9d)

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore,

a1 = [14 - 5(2)] / 2 = [14 - 10] / 2 = 4/2 = 2

And

a13 = [2 + 2(12)] = 26

Answer 2

☺HERE IS YOUR ANSWER☺

given -
⭐S6 = 42

⭐n/2( 2a + (n-1)d) = 42
2a+ 5d = 14------- (1)eq.

⭐now , according to qus.-
a+9d/ a+29d = 1/3
3a + 27d = a+ 29d
so, 2a = 2d ---(2)

⭐then by 1& 2--

7d= 14
so , d= 2
then , a = 2

hence , A13 = a +12d = 26 ans.

☺hope it helps u☺