WHAT AM I DOING WRONG?

[Maryanne]

o.k I know this is not a difficult problem but I keep making dumb mistakes and have not been able to get the answer! Any help would be appreciated!

Question: The first eight terms of an arithmetic series have a sum of 48. The common difference is 3. What are the first three terms of the series?


S8=48

cd=3

t1, t1+d, t1+2d,t1,3d, t1+4d,t1+5d,t1+6d,t1+7d

then i grouped togather like terms;

8t1+28d= 48

8t1+28(3)=48

t1=-4.5


Which doesn't work!

 

[pka]

In your 8t1+28d= 48 .
8t1= -15

 

[galactus]

Solve

\(\displaystyle \L\\\frac{n}{2}[2a_{1}+(n-1)d]=48\ for\ a_{1}\)

Go from there.

 

[Maryanne]

where did u get the 21 from?

 

[TchrWill]

o.k I know this is not a difficult problem but I keep making dumb mistakes and have not been able to get the answer! Any help would be appreciated!

Question: The first eight terms of an arithmetic series have a sum of 48. The common difference is 3. What are the first three terms of the series?


S8=48

cd=3

t1, t1+d, t1+2d,t1,3d, t1+4d,t1+5d,t1+6d,t1+7d

then i grouped togather like terms;

8t1+28d= 48

8t1+28(3)=48

t1=-4.5


Which doesn't work

The sum of an arithmetic progression is S = n(a + L)/2 where a = the first term. L = the last term and n = the number of terms.

We then have 49 = 8(L + a)/2 making (L + a) = 12.

The last term of such a progression is L = a + (n - 1)d where d = the common difference.

We then have (L - a) = 7(3) = 21

Adding the two expressions, we get 2L = 33 making L = 16.5.

Working backwards, the first term is - 4.5. making the sequence -4.5, -1,5, 1.5, 4.5, 7.5, 10.5, 13.5, and 16.5 allof which sum to 48.

So you were right in the first place.