Given the ratio of the sum of $n$ terms of two arithmetic progressions, find the ratio of their $m$-th terms

Question

The ratio of sum of $n$ terms of two arithmetic progressions is $$7n + 1 : 4n +27$$ Find the ratio of their $m$-th terms.

I tried everything but couldn't get the answer. Here is what I tried:-

1. $\dfrac{\frac{n}{2}(2a+(n-1)d)}{\frac{n}{2}(2a +(n-1)d)}$
2. $\dfrac{2a + (n-1)d}{2a + (n-1)d}$

And $m$ terms are $a+(m-1)d$.

I tried to simplify, expand but couldn't get the answer.

Answer 1

The ratio of the sum =$$\dfrac{\dfrac n2\{2a_1+(n-1)d_1\}}{\dfrac n2\{2a_2+(n-1)d_2\}}=\dfrac{a_1+\dfrac{n-1}2\cdot d_1}{a_2+\dfrac{n-1}2\cdot d_2}$$

Set $\dfrac{n-1}2=m-1\iff n=?$

We know the $m$th term is $$a_1+(m-1)d_1$$ where $a_1,d_1$ are the first term & the common difference respectively.

Answer 2

$$\frac{\sum a_n}{\sum b_n}=\frac{A_n}{B_n}=\frac{7n+1}{4n+27}$$ Where $A_n$ and $B_n$ are sum upto $n$ terms in the respective series. This implies $$A_n=kn(7n+1)\hspace{1cm} \mathrm{and}\hspace{1cm}B_n=kn(4n+27)$$ Now we can calculate $a_m$ which is equal to $$A_m-A_{m-1} = km(7m+1)-k(m-1)(7(m-1)+1)=k(14m-6)$$ and $b_m$ is $$B_m-B_{m-1}=km(4m+27)-k(m-1)(4(m-1)+27)=k(8m+23)$$ So, $$\frac{a_m}{b_m}=\frac{14m-6}{8m+23}$$