The sum of first 6 terms of an AP is 42. The ratio of its 10th term to its 30th term is 1:3 . Calculate the first and the thirteenth term of the AP.

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Solution

As usual, Let the first term be a and the common difference be d

thus, A to Q,

S₆ = 42

=> 6/2 (2a + 5d) = 42

=> 3 (2a + 5d) = 42

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

Again, T₁₀/T₃₀ = 1 : 3

=> (a + 9d)/(a + 29d) = 1/3

=>a + 29d = 3a + 27d

=> 2a - 2d = 0

=> a - d = 0 --------- (ii)

On solving (i) and (ii) we get,

a = 2 and d = 2

Thus, first term = 2 and

Thirteenth term = 2 + 12(2) = 2 + 24 = 26

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