**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, N_{11} = a + (10 – 1) d = a + 9 d

and 30^{th} term, N_{30 }= 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, S_{6} = (2 a + (6 – 1) d) = 3 (2 a + 5 d)

It is given that Sum of first 6 terms, S_{6} = 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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