Question 1195430
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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<br>
{{{(3+x)/2=8}}}
{{{3+x=16}}}
{{{x=13}}}<br>
So the 6th term is 13.<br>
That 6th term, 13, is the first term, 3, plus the common difference, d, 5 times:<br>
{{{13=3+5d}}}
{{{10=5d}}}
{{{d=2}}}<br>
So the AP has first term 3 and common difference 2.<br>
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:<br>
{{{3+(n-1)2=3+2n-2=2n+1}}}<br>
The sum of all the terms, 168, is the number of terms, n, times the average of the first and last terms:<br>
{{{S=n(3+(2n+1))/2=n(n+2)=168}}}
{{{n^2+2n=168}}}
{{{n^2+2n-168=0}}}
{{{(n+14)(n-12)=0}}}
{{{n=12}}}<br>
The number of terms in the whole sequence is 12.<br>
The last (12th) term is 3+11(2)=25.<br>
ANSWERS:
common difference: 2
number of terms: 12
last term: 25<br>
CHECK:
(Sum = number of terms times average of first and last)
12((3+25)/2)=12(28/2)=12*14=168<br>