# Arithmetic progression sum and nth term

1. Feb 17, 2015

### Suraj M

1. The problem statement, all variables and given/known data
The ratio of sums of 2 AP for n terms each is $\frac{3n + 8}{7n + 15}$
that is
$${\frac{s_a}{s_b}} = \frac{3n + 8}{7n + 15}$$
find the ratio of their 12th terms.
$$Required= \frac{a₁_a+(n-1)d_a}{a_b + (n-1)d_b}$$

2. Relevant equations
Tn = a + (n-1)d

3. The attempt at a solution
OKay the regular way that gives the right answer is this..
$${\frac{2a_a + (n-1)d_a}{2a_b + (n-1)d_b} } = {\frac{3n + 8}{7n + 15}} ~~ eq1 ~~$$
$$required = \frac{a_a+11d_a}{a_b+11d_b}$$
so putting n = 23 in the eq 1 we get the answer as $\frac{7}{16}$
but i tried another method..
$${\frac{s_a}{s_b}} ={ \frac{3n + 8}{7n + 15}}$$
here we can say..
$$s_a = (3n + 8)x$$
$$s_b = (7n + 15)x$$
if i put n = 1 ,2 ,3 for both these AP's
for AP 1: using sa
s₁=a₁ right??
then a₁=11x
then s₂=14x so a₂ = s2 -s1 = 3x
then s3 =17x so a3 = s3-s2 = 3x
how is a2=a3 ??
similarly even for AP2
but why??

Last edited: Feb 17, 2015
2. Feb 17, 2015

### Svein

deleted...

Last edited: Feb 17, 2015
3. Feb 17, 2015

### Svein

OK. Let AP1 be given as $a_{n}$ and AP2 be given as $b_{n}$ Then the sums are given as $s_{a,n}=\frac{n(a_{1}+a_{n})}{2}$ and $s_{b,n}=\frac{n(b_{1}+b_{n})}{2}$. The ratio is therefore $\frac{s_{b,n}}{s_{a,n}}=\frac{(b_{1}+b_{n})}{(a_{1}+a_{n})}$. Substitute the expression for $a_{n}$ and $b_{n}$: $\frac{s_{b,n}}{s_{a,n}}=\frac{(b_{1}+b_{1}(n-1)e)}{(a_{1}+a_{1}(n-1)d)}=\frac{3n+8}{7n+16}$. Solve for $\frac{b_{1}}{a_{1}}$ and insert...

4. Feb 17, 2015

### Suraj M

No Svein, i know how to get the answer, my question is why am i getting the terms after the first term, equal by the method I've shown above?

5. Feb 17, 2015

### HallsofIvy

I think you are confusing yourself with your notation. You say
Initially, you wrote that s1 and s2 were two different arithmetic progressions. Now you are using "s1" and "s2" to mean the first two partial sum of one arithmetic progression.

6. Feb 17, 2015

### Suraj M

Ok fine, just notation mistake. check the original post, now ok?
let those s1 s2 s3 be fore the first AP only.
then a2=a3=... why?

7. Feb 24, 2015

### Suraj M

Hello? anyone?

8. Feb 25, 2015

### haruspex

You are assuming x is a constant, independent of n. It need not be. It is only necessary that it is the same function of n in nu numerator and denominator.

9. Feb 25, 2015

### Suraj M

x is the common factor, it must be constant right?

10. Feb 25, 2015

### haruspex

Why might it not be, say, n? I.e., sa = (3n+8)n, etc.

11. Feb 26, 2015

### Suraj M

ohh, did not not think of that, sorry, but then the common factor may be any function of n. Is there no way of finding that common factor between sa and sb?

12. Feb 26, 2015

### haruspex

You know that sa is the sum of an AP, so it must be a quadratic function of n. You also know one of its factors.