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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.

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  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Sep 25 '17 at 9:46
  • $\begingroup$ I cleaned-up the typesetting, but you should be clearer about what you mean. As written, each of the fractions you provided simply reduce to $1$. Presumably, the "$a$"s and "$d$"s should be different in the numerators and denominators to represent that they come from separate arithmetic progressions. $\endgroup$ – Blue Sep 25 '17 at 9:49
  • $\begingroup$ @Shaun Yes thank you.I need to learn that. $\endgroup$ – Ram Keswani Sep 25 '17 at 10:11
  • $\begingroup$ @Blue yes sir sorry I will learn mathjax now. $\endgroup$ – Ram Keswani Sep 25 '17 at 10:11
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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.

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  • $\begingroup$ Sir why n-1/2 = m-1? $\endgroup$ – Ram Keswani Sep 25 '17 at 10:14
  • $\begingroup$ @RamKeswani, As we need to match with $$a_i+(m-1)d_i$$ $\endgroup$ – lab bhattacharjee Sep 25 '17 at 10:49
  • $\begingroup$ sorry sir I dont understand can you provide the complete answer. $\endgroup$ – Ram Keswani Sep 25 '17 at 11:22
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$$\frac{\sum a_n}{\sum b_n}=\frac{A_n}{B_n}=\frac{7n+1}{4n+27}$$ 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, $$\frac{a_m}{b_m}=\frac{14m-6}{8m+23}$$

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  • $\begingroup$ why An = kn(7n+1) Shouldnt it just be k (7n+1)? $\endgroup$ – Ram Keswani Sep 25 '17 at 10:18
  • $\begingroup$ Because then you would have to use a different k for $A_{n-1}$ $\endgroup$ – Piyush Divyanakar Sep 25 '17 at 10:48
  • $\begingroup$ I tried using what you have suggested it gives the answer 7/135 which is incorrect. The answer I have provided does add up to the sum ratio that you have mentioned. $\endgroup$ – Piyush Divyanakar Sep 25 '17 at 10:49

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