Algebra in sequence

[mattgad]

The fourth term of an arithmetic sequence is 3k, where k is a constant, and the sum of the first 6 terms of the series is 7k +9.

Show that the first term of the series is 9 - 8k.

So far, I've got this:

U4 = 3k
U1 + U2 + .... + U9 = 7k + 9

By definition:

U2 = U1 + d
U3 = U1 + 2d
U4 = U1 + 3d
And so on till..
U9 = U1 + 8d

So,

U4 = U1 + 3d and U4 = 3k

So, 3k = U1 + 3d, U1 = 3k - 3d

6U1 + 15d = 7k + 9

Subing in U1

6(3k - 3d) + 15d = 7k + 9, 18k - 18d + 15d = 7k + 9

What do I need to do next?

Thanks.

 

[Unco]

Impressive work.

You can solve

18k - 18d + 15d = 7k + 9

for -3d

and substitute this into

U1 = 3k - 3d

to find U1.

 

[mattgad]

I've now solved that to give me the correct answer.

b) Find an expression for the common difference in terms of k

U4 = U1 + 3d, so

U4 = 9 - 8k + 3d
3k = 9 - 8k + 3d

d = 11k - 9 divided by 3

Correct.

c) Given U7 = 12, Find value of k:

U7 = U1 + 6d
U7 = 9 - 8k + 22k - 18 = 14k - 9

12 = 14k - 9
14k = 21
k = 1.5

Answer is 1.5, correct.

d) Find sum of first 20 terms.

k = 1.5 so d = -17.5/3 = 5.833333333333

a = 9 - 8k = 9 - 8(1.5) = -3

Sn = n/2(2a + d(n - 1))
Sn = 20/2(-6 + 5.8333333(19)) = 1058.3333

Answer is 415, where did I go wrong?

 

[pka]

I do not think that it is true as stated!
Lets assume the answer, that is U<SUB>1</SUB>=9−8k.
Then U<SUB>4</SUB>= U<SUB>1</SUB>+3d=(9−8k)+3d=3k.
From this we get d=4k−3.
The sum of the first six terms is 6U<SUB>1</SUB>+15d=6(9−8k)+15(4k−3)=12k−9.

 

[mattgad]

The book answer also tells me d = 11k - 9 divide 3

3k = 9 - 8k + 3d
3d = 3k + 8k - 9
3d = 11k - 9
d = 11k - 9 divide 3

I think somehow you may have ended up with 3d = 12k - 9?

 

[pka]

Sorry, I can not add!

 

[mattgad]

We all make silly mistakes like that.

U20 = 20U1 + 190d


EDIT:

Noticed a mistake, got d = 5.5, so U1 = -35

20(-35) + 190(5.5) = 345, 70 off the answer. Can you spot a mistake I cant?

 

[galactus]

\(\displaystyle d=\frac{11k-9}{3}\), k=1.5

\(\displaystyle d=\frac{11(1.5)-9}{3}=2.5\)

\(\displaystyle 10(-6+(2.5)19)=415.\)

Solving the system would have given the 3 results we need:

\(\displaystyle \begin{array}{cc}1&3&-3&0\\1&6&0&12\\6&15&-7&9\end{array}\)

\(\displaystyle =\begin{array}{cc}1&0&0&-3\\0&1&0&5/2\\0&0&1&3/2\end{array}\)

\(\displaystyle U=-3\)
\(\displaystyle d=2.5\)
\(\displaystyle k=1.5\)

 

[mattgad]

I don't see how that adds up to 415.

10(-6 + 22.5) = 10(16.5) = 165?

My mistake, I was x by 9, not 19. Sorry.

 

[pka]

In the clear morning light!

\(\displaystyle U_4 = U_1 + 3d = 3k\; \Rightarrow U_1 = 3k - 3d\)

The sum of the first six: \(\displaystyle \sum\limits_{k = 1}^6 {\left[ {U_1 + \left( {k - 1} \right)d} \right]} = 6U_1 + 15d = 6\left( {3k - 3d} \right) + 15d = 18k - 3d\)

Thus: \(\displaystyle 18k - 3d = 7k + 9\quad \Rightarrow \;d = \frac{{11k - 9}}{3}\).

From which we get, \(\displaystyle U_1 = 3k - 3d = 3k - (11k - 9) = 9 - 8k.\)

 

[Gene]

Or we have 6U1+15d=7k+9 or
6U1+15d-7k=9

U1+3d=3k or
u1+3d-3k=0 or
5U1+15d-15k = 0
Subtracting
U1+8k=9
QED

That's probably the same as above, but I have trouble reading things on WebTV.