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The sum of the first six terms of an arithmetic progression (AP) is 42. The ratio of its 10th term to its 30th term is 1:3. Calculate the first and the 13th term of the AP.

Answer :

Final answer:

The question involves solving for the first and thirteenth terms of an Arithmetic Progression (AP) with the known sum of the first six terms and a given ratio of the 10th to the 30th term. We find the first term 'a' is 1 and the common difference 'd' is also 1, thus making the 13th term equal to 13.

Explanation:

The question involves finding the first and thirteenth term of an Arithmetic Progression (AP) given two conditions:

  • The sum of the first six terms of the AP is 42.
  • The ratio of the 10th to the 30th term is 1:3.

To solve this, we can use the formulas for the nth term and the sum of the first n terms of an AP. The nth term of an AP is given by Tn = a + (n-1)d, where a is the first term and d is the common difference. The sum of the first n terms of an AP is given by

[tex]Sn = (n/2)(2a + (n-1)d)[/tex].

Using these formulas and the given information, we can set up a system of equations:

(1) 42 = (6/2)(2a + (6-1)d)

(2) [tex]T10/T30 = (a + 9d)/(a + 29d) = 1/3[/tex]

By simplifying equation (1), we get:

42 = 3(2a + 5d) => 42 = 6a + 15d => 7 = 2a + 5d

From equation (2), we can write:

(a + 9d) = ⅓(a + 29d)

By cross-multiplying and solving, we get:

3a + 27d = a + 29d => 2a = 2d => a = d

Substituting a = d into the first equation, we find:

7 = 2a + 5a => 7 = 7a => a = 1

Now, to find the 13th term (T13), we use the nth term formula:

[tex]T13 = a + (13-1)d = > T13 = 1 + 12(1) = > T13 = 13[/tex]

So, the first term of the AP is 1, and the thirteenth term is 13.

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