Arithmetic progression sum and nth term

[Suraj M]

Homework Statement

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.
$$ Required= \frac{a₁_a+(n-1)d_a}{a_b + (n-1)d_b}$$

Homework Equations

T_n = a + (n-1)d

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 s_a
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??

 

[Svein]

deleted...

 

[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...

 

[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?

here we can say..

s₁=(3n+8)x​

s₁ = (3n + 8)x

s₂=(7n+15)x​

s₂ = (7n + 15)x
if i put n = 1 ,2 ,3 for both these AP's
for AP 1:
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 ??

 

[HallsofIvy]

I think you are confusing yourself with your notation. You say

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 ??

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.

 

[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?

 

[Suraj M]

Hello? anyone?

 

[haruspex]

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?

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.

 

[Suraj M]

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

 

[haruspex]

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

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

 

[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 s_a and s_b?

 

[haruspex]

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 s_a and s_b?

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