The sun of the first 6 term of ap is 42 the ratio of 10 term to 30 term is 1:3 find the 15 term
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Let a1 = the first term....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
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