# Arithmetic progression

1. Oct 9, 2008

### Lasher

1. The problem statement, all variables and given/known data

In an arithmetic progression, the sum of the first 10 terms is the same as the sum of the next 5 terms. Given that the first term is 12, find the sum of the first 15 terms.

2.
The only one I could think of is
S= n/2 (2a+(n-1)d)

3.
I've tried solving it, but failed. I guess this is due to lack of experience, as I missed the topic.

2. Oct 9, 2008

### Mentallic

That is the correct equation to use in this case, but some ingenuity is needed to apply it.

To find the sum of the first n terms in an Arithmetic Progression, use: $$S_{n}=\frac{n}{2}[2a+(n-1)d]$$

Now it tells you that the sum of the first 10 terms, i.e. S10 = The sum of the next 5 terms, i.e. T11 + T12 +...+ T15.

Now, $$S_{10}=T_{1}+T_{2}+...+T_{10}$$

and, $$S_{15}=T_{1}+T_{2}+...+T_{15}$$

Therefore, $$T_{11}+T_{12}+...+T_{15}=S_{15}-S_{10}$$

From here use the formula to re-arrange and simplify to find the difference (d).

Once you find the difference, use the formula again to find S15

3. Oct 9, 2008

### Lasher

Ooph.. I get the first part, but now I have trouble finding d.
All I get is d=S5-6
Would you be so kind to explain step by step how to find d?

4. Oct 9, 2008

### Mentallic

Homework helper rules state that I can't provide complete solutions, but don't worry, that doesn't stop you from finding the answer.

ok so from summarizing the question and realizing how it needs to be expressed in an equation we get:

$$S_{10}=T_{11}+T_{12}+...+T_{15}$$
$$S_{10}=S_{15}-S_{10}$$ (1)
Now use the formula and substitute ~ $$S_{n}=\frac{n}{2}[2a+(n-1)d]$$ for all in (1)
$$a=12$$ (first term of sequence - given)

5. Oct 11, 2008

### Lasher

Thank you so much))

Just to be sure, is the answer S15=240 ?

6. Oct 11, 2008

### Mentallic

Well lets check shall we?

$$S_n=\frac{n}{2}[2a+(n-1)d]$$, Where n=15, S15=240, a=12

Therefore, $$240=\frac{15}{2}[2.12+(15-1)d]$$

$$480=15(24+14d)$$ (multiply through by 2 and simplify)
$$480=30(12+7d)$$ (common multiple of 2 in factor)
$$16=12+7d$$ (divide by 30)
$$d=\frac{4}{7}$$

You may need to check over your work since this is not the difference I got by my calculations (and I've double-checked). Depending on what you calculated for d, depends on where you might have gone wrong.

Are you sure you substituted $$S_n=\frac{n}{2}[2a+(n-1)d]$$ correctly into $$S_{10}=S_{15}-S_{10}$$?

7. Oct 11, 2008

### Lasher

Seems like math is my worst enemy)

What I've done is

$$\frac{10}{2}[2(12)+(10-1)d]=\frac{15}{2}[2(12)+(15-1)d] - \frac{10}{2}[2(12)+(10-1)d]$$

8. Oct 11, 2008

### Mentallic

Yes you've set it out correctly. What you should find is d=4, if you are still not getting this answer I'll post a step-by-step for you

9. Oct 11, 2008

### Lasher

Ok..so, there I go
$$\frac{10}{2}[2(12)+(10-1)d]=\frac{15}{2}[2(12)+(15-1)d] - \frac{10}{2}[2(12)+(10-1)d]$$

Therefore $$5[24+9d]=7.5[24+14d] - 5[24+9d]$$

Thus $$120+45d=180+105d-120+45d$$

So, $$45d=180-120+105d+45d-120$$

$$45d=60-120+150d$$

$$-105d=-60$$

$$d=-60\div-105$$

$$d=\frac{4}{7}$$

There o__O

10. Oct 11, 2008

### Mentallic

Yes I thought that is where you might have gone wrong. To save yourself a lot of effort, you could have noticed that $$S_{10}=S_{15}-S_{10}$$ is much easier to solve if simplified to $$2S_{10}=S_{15}$$, but it doesn't make much of a difference.

$$5[24+9d]=7.5[24+14d]- 5[24+9d]$$

The part at the end $$-5[24+9d]$$ is expanded to $$-120-45d$$
not $$-120+45d$$

The negative 5 outside the factor is multiplied by each term inside the factor. Do not forget those negatives

11. Oct 11, 2008

### Lasher

Oh, that's why))) Thank you))))