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Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.

My method:

Given:

$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$

$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$

$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\left(n-1\right)+8\right)\right)}{k\left(\left(7n+15\right)-\left(7\left(n-1\right)+15\right)\right)}=\frac{3}{7}$

As this applies for any term:

$\frac{T_{12}}{t_{12}}=\frac{3}{7}$

But this is not the answer. The actual answer is $\frac7{16}$. I know how to obtain that answer.

But why is my solution wrong? Its probably the concept I guess.

hrkrshnn
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Hijaz Aslam
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5 Answers5

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If what you call $k$ is independent of $n$, then your argument works fine. But there is no reason to assume independence.

André Nicolas
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  • So can we argue independence if the A.Ps considered here are the same but the terms are different for instance ($m^{th}$ and $n^{th}$ term? ) – Hijaz Aslam Aug 11 '14 at 15:10
  • I do not understand the question in the comment well enough to answer. – André Nicolas Aug 11 '14 at 15:14
  • For example in the questions:

    If the ratio of the sum of m terms and n terms of an A.P be $m^2:n^2$, prove that the ratio of its mth and nth term will be 2m-1:2n-1.

    This method works fine for this question.

     – Hijaz Aslam Aug 11 '14 at 15:16
  • It just so happens that in that case the condition forces the sums to be $km^2$, $kn^2$ where $k$ is really constant. Set up the equality, do a bit of algebra, there is nice cancellation. – André Nicolas Aug 11 '14 at 16:01
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$${S_n\over3n+8}={s_n\over7n+15},$$ but we cannot know that $${S_n\over3n+8}={s_n\over7n+15}\overset{?}{=}{S_{n-1}\over3(n-1)+8}={s_{n-1}\over7(n-1)+15}.$$ Actually, $${S_n\over3n+8}={s_n\over7n+15}=k_\color{red}{n}.$$

Jaehyeon Seo
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Hint.

We have that $$ S_n=A_1+\cdots+A_n, $$ where $A_n=Kn+L$, and hence $$ S_n=K\frac{n(n+1)}{2}+Ln. $$ Similarly, $s_n=a_1+\cdots+a_n$, where $a_n=kn+\ell$ and $s_n=k\frac{n(n+1)}{2}+\ell n$, and $$ \frac{S_n}{s_n}=\frac{K\frac{n(n+1)}{2}+Ln}{k\frac{n(n+1)}{2}+\ell n}=\frac{K(n+1)+2L}{k(n+1)+2\ell}=\frac{Kn+2L+K}{kn+2\ell+k}=\frac{3n+8}{7n+15}, $$ and $$ \frac{A_{12}}{B_{12}}=\frac{12K+L}{12k+\ell} $$

Yiorgos S. Smyrlis
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The trick is to notice that we can safely cancel constant common factors (if any), and stick in the $n$ in both numerator and denominator:

$\dfrac{T_n}{t_n} = \dfrac{S_n - S_{n - 1}}{s_n - s_{n - 1}} = \dfrac{n(3n + 8) - (n - 1)(3(n - 1) + 8)}{n(7n + 15) - (n - 1)(7(n - 1) + 15)} = \dfrac{6n + 5}{14n + 8}$

Plug in $n = 12$ to get the ratio of 12th terms.

AgentS
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A short and direct answer to your question, distilling the wisdom in answers given by others here earlier:

Both $S_n$ and $s_n$ are of the form $n(An+B)$.

However when you take the ratio, $n$ gets cancelled out.

Therefore when you take the ratio of the $\underline{\text{difference}}$, $n$ (and also $n-1$) should be reinstated accordingly to arrive at the correct answer.

In summary,

$$\begin{align} \dfrac{S_n}{s_n}&=\dfrac{3n+8}{7n+15}=\dfrac{n(3n+8)}{n(7n+15)}\\ \dfrac{S_{12}}{s_{12}}&=\dfrac {44}{99}\qquad =\dfrac {12\cdot 44}{12\cdot 49}\\ \dfrac{S_{12}-S_{11}}{s_{12}-s_{11}}&\quad\qquad\;\quad=\dfrac{12\cdot44-11\cdot41}{12\cdot 99-11\cdot 92}=\dfrac{48-41}{108-92}=\dfrac7{16}\\ \end{align}$$

Hypergeometricx
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