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

Guest Feb 23, 2015

#1**+5 **

Let a_{1} = the first term....we have....the sum of the 1st 6 terms = 42....therefore...

42 = (6/2)[2a_{1} + 5d]

42 = 3[2a_{1} + 5d]

14 = 2a_{1} + 5d → a_{1} = [14 - 5d]/2

And a_{30} = 3(a_{10}) .... therefore.....

3(a_{10}) = a_{10} + 20d

2(a_{10}) = 20d

a_{10} = 10d

So we have

a_{10} = (a_{1} + 9d) ... and by substitution.....

10d = ([14- 5d]/2 + 9d)

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a_{1} = [14 - 5(2)] / 2 = [14 - 10] / 2 = 4/2 = 2

And

a_{13} = [2 + 2(12)] = 26

CPhill
Feb 23, 2015

#1**+5 **

Best Answer

Let a_{1} = the first term....we have....the sum of the 1st 6 terms = 42....therefore...

42 = (6/2)[2a_{1} + 5d]

42 = 3[2a_{1} + 5d]

14 = 2a_{1} + 5d → a_{1} = [14 - 5d]/2

And a_{30} = 3(a_{10}) .... therefore.....

3(a_{10}) = a_{10} + 20d

2(a_{10}) = 20d

a_{10} = 10d

So we have

a_{10} = (a_{1} + 9d) ... and by substitution.....

10d = ([14- 5d]/2 + 9d)

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a_{1} = [14 - 5(2)] / 2 = [14 - 10] / 2 = 4/2 = 2

And

a_{13} = [2 + 2(12)] = 26

CPhill
Feb 23, 2015