Question

Please help me!
The ratio of sum of n terms of two A.P's is $(7n + 1):(4n+27)$. Find the ratio of their $m^{th}$ term.
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Answer 1

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• If the ratio of the sum of the first n terms of two AP is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.

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This is a very good question, just don't get complicated or confused.

Ratio of sum of 1st n terms of 2 A.P is (7n + 1) : (4n + 27).

Ratio of mth term?

$\sf Let \: the\: 1st \:terms \:of \:2\: A.P's\: be\: a_{1} \: and \: a'_{1}.$

$\sf Let \: the\: common\: differences\: \:2\: A.P's\: be\: d \: and \: d'.$

$\sf Let \: the\: sums \:of \:2\: A.P's\: be\: S_{1} \: and \: S'_{1}.$

$\sf Let \: the\: 2 \: A.P's\: be\: T_{1} \: and \: T'_{1}.$

According to the question,

• $\boxed{\pink{\sf \dfrac{T_{1}}{T'_{1}} \: = \: \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{\dfrac{n}{2} (2a \: + \: (n \: - \: 1) \: d}{\dfrac{n}{2} (2a' \: + \: (n \: - \: 1) \: d'}}}$

• $\sf \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{\cancel{\dfrac{n}{2}} (2a \: + \: (n \: - \: 1) \: d}{\cancel{\dfrac{n}{2}} (2a' \: + \: (n \: - \: 1) \: d'}$

• $\sf \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{2a \: + \: (n \: - \: 1) \: d}{2a' \: + \: (n \: - \: 1) \: d'}$

Now, we must Substituting (2m - 1) in place of n,

Substituting,

• $\sf \dfrac{7(2m \: - \: 1) \: + \: 1}{4(2m \: - \: 1) \: + \: 27} \: = \: \dfrac{2a \: + \: ((2m\: - \: 1) \: - \: 1) \: d}{2a' \: + \: ((2m \: - \: 1) \: - \: 1) \: d'}$

• $\sf \dfrac{7(2m \: - \: 1) \: + \: 1}{4(2m \: - \: 1) \: + \: 27} \: = \: \dfrac{2a \: + \: (2m \: - \: 2) \: d}{2a' \: + \: (2m \: - \: 2) \: d'}$

• $\sf \dfrac{14m \: - \: 7 \: + \: 1}{8m \: - \: 4 \: + \: 27} \: = \: \dfrac{2a \: + \: (2m \: - \: 2) \: d}{2a' \: + \: (2m \: - \: 2) \: d'}$

• $\sf \dfrac{14m \: - \: 6}{8m \: + \: 23} \: = \: \dfrac{2a \: + \: (2m \: - \: 2) \: d}{2a' \: + \: (2m \: - \: 2) \: d'}$

Taking 2 common on left side,

• $\sf \dfrac{14m \: - \: 6}{8m \: + \: 23} \: = \: \dfrac{2(a \: + \: (m \: - \: 1) \: d)}{2(a' \: + \: (m \: - \: 1) \: d')}$

• $\sf \dfrac{14m \: - \: 6}{8m \: + \: 23} \: = \: \dfrac{\cancel{2}(a \: + \: (m \: - \: 1) \: d)}{\cancel{2}(a' \: + \: (m \: - \: 1) \: d')}$

• $\sf \dfrac{14m \: - \: 6}{8m \: + \: 23} \: = \: \dfrac{a \: + \: (m \: - \: 1) \: d}{a' \: + \: (m \: - \: 1) \: d'}$

• $\sf \dfrac{14m \: - \: 6}{8m \: + \: 23} \: = \: \dfrac{a_m}{a'_m}$

• $\boxed{\sf a_m : a'_m \: = \: 14m \: - \: 6 : 8m \: + \: 23}$

Therefore,

• Ratio of mth term is 14m - 6 : 8m + 23.

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Why should we substitute (2m - 1)?

• We should substitute (2m - 1) so as to take 2 common from the RHS and get the mth term i.e $\sf a_m$ and get the ratio.

Answer 2

Required Answer :-

According to the question

$\sf \dfrac{Sn}{Sn'} =\dfrac{ \dfrac{n}{2}\bigg(2a + [n-1] d\bigg)}{\dfrac{n}{2}\bigg(2a'+[n-1]d'\bigg)}$

$\sf\dfrac{Sn}{Sn'} = \dfrac{\bigg(2a\bigg)[n-1]d}{\bigg(2a'\bigg)+[n-1]d'}$

$\sf\dfrac{Sn}{Sn'} = \dfrac{2a+[n-1]d}{2a'+[n-1]d'}$

$\sf \dfrac{2a+ [n - 1]d}{2a' + [n - 1]d'}=\dfrac{7n + 1}{4n+27}$

$\sf Exchanging\; n =\; 2m-1$

$\sf \dfrac{7(2m-1)+1}{4(2m-1)+27}$

$\sf \dfrac{14m-7+1}{8m - 4 + 27}$

$\sf\dfrac{14m-6}{8m+23}$