SOLUTION: The first term of an AP is 3. Given that the sum of the first 6 terms is 48 and that the sum of all the terms is 168 .calculate the common difference,the number of terms in the AP

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Question 1195430: The first term of an AP is 3. Given that the sum of the first 6 terms is 48 and that the sum of all the terms is 168 .calculate the common difference,the number of terms in the AP and the last term
Found 3 solutions by greenestamps, ikleyn, MathTherapy:
Answer by greenestamps(13200) About Me  (Show Source):
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The sum of an AP is the number of terms, multiplied by the average of the first and last terms. Given first term 3 and sum the first six terms 48, the average is 48/6 = 8. If the 6th term is x, then

%283%2Bx%29%2F2=8
3%2Bx=16
x=13

So the 6th term is 13.

That 6th term, 13, is the first term, 3, plus the common difference, d, 5 times:

13=3%2B5d
10=5d
d=2

So the AP has first term 3 and common difference 2.

Now let n be the number of terms in the whole sequence. The last term is the first term, 3, plus the common difference, d, (n-1) times:

3%2B%28n-1%292=3%2B2n-2=2n%2B1

The sum of all the terms, 168, is the number of terms, n, times the average of the first and last terms:

S=n%283%2B%282n%2B1%29%29%2F2=n%28n%2B2%29=168
n%5E2%2B2n=168
n%5E2%2B2n-168=0
%28n%2B14%29%28n-12%29=0
n=12

The number of terms in the whole sequence is 12.

The last (12th) term is 3+11(2)=25.

ANSWERS:
common difference: 2
number of terms: 12
last term: 25

CHECK:
(Sum = number of terms times average of first and last)
12((3+25)/2)=12(28/2)=12*14=168

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
The first term of an AP is 3. Given that the sum of the first 6 terms is 48
and that the sum of all the terms is 168, calculate the common difference,
the number of terms in the AP and the last term
~~~~~~~~~~~~~~~~~~~~ 

The sum of the first 6 terms is 48.


This sum is 6 times the average of the first 6 terms

and is 6 times the average of the 1st and 6th terms of this AP

    6%2A%28%28a%5B1%5D%2Ba%5B6%5D%29%2F2%29 = 48,

or

    6%2A%283+%2B+a%5B6%5D%29 = 48*2

    18+%2B+6%2Aa%5B6%5D = 96

    6%2Aa%5B6%5D = 96 - 18 = 78

    a%5B6%5D = 78/6 = 13.


Thus  the 6th term of the AP is 13:  

    a%5B6%5D = a%5B1%5D%2B5%2Ad = 13

    3 + 5d = 13

        5d = 13-3 = 10

         d = 10/5 = 2.


Thus the common difference of the AP is  d= 2.

    
Next, the sum of all the terms of the AP is 168

    S%5Bn%5D = %282a%5B1%5D+%2B+%28n-1%29d%29%2A%28n%2F2%29 = 168

or

    (2*3+(n-1)*2)*n = 168*2

    6n + 2*(n-1)*n = 336

    2n^2 + 4n - 336 = 0

     n^2 + 2n - 168 = 0

     (n-12)*(n+14) = 0


The roots of this quadratic equation are 12 and -14; since we are looking for the number of terms n,
we see that the solution is the positive root n= 12.


ANSWER.  The common difference of the AP is 2;

         the number of terms is 12, and the last term is  a%5B12%5D = 3 + 2*(12-1) = 25.

Solved.

All questions are answered.

------------------

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Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The first term of an AP is 3. Given that the sum of the first 6 terms is 48 and that the sum of all the terms is 168 .calculate the common difference,the number of terms in the AP and the last term 
                                          Sum of an AP: matrix%281%2C3%2C+S%5Bn%5D%2C+%22=%22%2C+%28n%2F2%29%282a%5B1%5D+%2B+%28n+-+1%29d%29%29
Sum of this AP with its first 6 terms summing to 48: 
                                   Common difference, or 

                                          Sum of an AP: matrix%281%2C3%2C+S%5Bn%5D%2C+%22=%22%2C+%28n%2F2%29%282a%5B1%5D+%2B+%28n+-+1%29d%29%29
       Sum of this AP with its terms summing to 168: 
                                                 (n - 12)(n + 14) = 0
             Number of terms in series, or  n = 12  or  n = - 14 (ignore) 

                                       Specific term of an AP: matrix%281%2C3%2C+a%5Bn%5D%2C+%22=%22%2C+a%5B1%5D+%2B+%28n+-+1%29d%29
                                 Last term (12th) of this AP: 
                       Last term (12th) of this AP, or  a12 = 25