Q) The ratio of the 10th term to its 30th term of an A.P. is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference of A.P.

Ans:

Step 1: We know that n^th term of an A.P.  =  a + (n-1) d

Therefore,  10^th term, N₁₁ = a + (10 – 1) d = a + 9 d

and 30^th term, N₃₀ = a + (30 – 1) d = a + 29 d

Given that : =

    =

∴ 3(a + 9 d) = a + 29 d

∴ 3 a + 27 d = a + 29 d

∴ 2 a = 2 d

∴ a = d ………….. (i)

Step 2:

We know that sum of n terms of an A.P.  S_n = (2a + (n-1) d)

Sum of first 6 terms, S₆ = (2 a + (6 – 1) d) = 3 (2 a + 5 d)

It is given that Sum of first 6 terms, S₆ =    42

∴ 3 (2 a + 5 d) = 42

From equation (i), we calculated a = d

∴ 3 (2 a + 5 a) = 42

∴ 21 a = 42

∴ a =

∴ a = 2

Since d = a            [from equation (i) ]

∴ d = 2

Therefore, first term of the AP is 2 and common difference is 2.

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