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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

Best Answer 

 #1
avatar+81051 
+5

Let a1  = the first term....we have....the sum of the 1st 6 terms  = 42....therefore...

42 = (6/2)[2a1 + 5d]

42 = 3[2a1 + 5d]

14 = 2a1 + 5d  →   a1 = [14 - 5d]/2 

And a30 = 3(a10)   .... therefore.....

3(a10) = a10 + 20d

2(a10) = 20d

a10 = 10d

So we have

a10 = (a1 + 9d)  ... and by substitution.....

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

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a1 = [14 - 5(2)] / 2  = [14 - 10] / 2 =  4/2 = 2

And

a13 = [2 + 2(12)] =  26

 

CPhill  Feb 23, 2015
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1+0 Answers

 #1
avatar+81051 
+5
Best Answer

Let a1  = the first term....we have....the sum of the 1st 6 terms  = 42....therefore...

42 = (6/2)[2a1 + 5d]

42 = 3[2a1 + 5d]

14 = 2a1 + 5d  →   a1 = [14 - 5d]/2 

And a30 = 3(a10)   .... therefore.....

3(a10) = a10 + 20d

2(a10) = 20d

a10 = 10d

So we have

a10 = (a1 + 9d)  ... and by substitution.....

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

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a1 = [14 - 5(2)] / 2  = [14 - 10] / 2 =  4/2 = 2

And

a13 = [2 + 2(12)] =  26

 

CPhill  Feb 23, 2015

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