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The sum of the first five terms of an arithmetic ...

The sum of the first five terms of an arithmetic sequence is 40 and the sum of the first ten terms is 155. Find an explicit formula for the nth term of the sequence.


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Love is the battery of life....

Hi, ------ S(5)= 40 = (a1+a5)*5/2= (2a1+4d)5a1+10d ---- 8=a1+2d ----- S(10)= 155 = (a1+a10)*10/2= (a1+a1+9d)*5 = 10a1+45d -------- 31=2a1+9d From the 1st equation: a1=8-2d install in the 2nd equation: 31=2(8-2d)+9d ----- 31=16-4d+9d ---- 15=5d d=3 and a1=2 ------- a9=a1+8d ------ a9=2+3*8=26 -------- Best regards,

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Ian
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a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d)

(a+5d) + (a+6d) + (a+7d) + (a+8d) + (a+9d)

5a + 10d = 40                                         (a)

10a + 45d = 155                                      (b)

10a + 20d = 80                                      2 x(a)

25d = 75                                           (b) - 2 x(a)

d = 3

a + 2d = 8 = a + 6                           from (a)/5

a = 2

Sequence is 2, 5, 8, etc.

nth term is a + (n -1)d = 2 + 3(n -1) = 3n -1

Every term is a multiple of three reduced by 1

Regards - Ian

 

 

 


 

Posted 2009-05-30T22:28:37Z
Ian was invited by Yedda to answer this question.

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