Please help me!
The ratio of sum of n terms of two A.P’s is [tex](7n + 1):(4n+27)[/tex]. Find the ratio of their [tex]m^{th}

2 thoughts on “Please help me!<br /> The ratio of sum of n terms of two A.P’s is [tex](7n + 1):(4n+27)[/tex]. Find the ratio of their [tex]m^{th}”

1. [tex] \huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Question:-}}}}}}} [/tex]

   • 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.

   [tex] \huge {\bf{\orange{\mathfrak{\dag{\underline{\underline{Answer:-}}}}}}} [/tex]

   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?

   [tex] \sf Let \: the\: 1st \:terms \:of \:2\: A.P’s\: be\: a_{1} \: and \: a’_{1}. [/tex]

   [tex] \sf Let \: the\: common\: differences\: \:2\: A.P’s\: be\: d \: and \: d’. [/tex]

   [tex] \sf Let \: the\: sums \:of \:2\: A.P’s\: be\: S_{1} \: and \: S’_{1}. [/tex]

   [tex] \sf Let \: the\: 2 \: A.P’s\: be\: T_{1} \: and \: T’_{1}. [/tex]

   According to the question,

   • [tex] \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’}}} [/tex]

   • [tex] \sf \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{\cancel{\dfrac{n}{2}} (2a \: + \: (n \: – \: 1) \: d}{\cancel{\dfrac{n}{2}} (2a’ \: + \: (n \: – \: 1) \: d’} [/tex]

   • [tex] \sf \dfrac{7n \: + \: 1}{4n \: + \: 27} \: = \: \dfrac{2a \: + \: (n \: – \: 1) \: d}{2a’ \: + \: (n \: – \: 1) \: d’} [/tex]

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

   Substituting,

   • [tex] \sf \dfrac{7(2m \: – \: 1) \: + \: 1}{4(2m \: – \: 1) \: + \: 27} \: = \: \dfrac{2a \: + \: ((2m\: – \: 1) \: – \: 1) \: d}{2a’ \: + \: ((2m \: – \: 1) \: – \: 1) \: d’} [/tex]

   • [tex] \sf \dfrac{7(2m \: – \: 1) \: + \: 1}{4(2m \: – \: 1) \: + \: 27} \: = \: \dfrac{2a \: + \: (2m \: – \: 2) \: d}{2a’ \: + \: (2m \: – \: 2) \: d’} [/tex]

   • [tex] \sf \dfrac{14m \: – \: 7 \: + \: 1}{8m \: – \: 4 \: + \: 27} \: = \: \dfrac{2a \: + \: (2m \: – \: 2) \: d}{2a’ \: + \: (2m \: – \: 2) \: d’} [/tex]

   • [tex] \sf \dfrac{14m \: – \: 6}{8m \: + \: 23} \: = \: \dfrac{2a \: + \: (2m \: – \: 2) \: d}{2a’ \: + \: (2m \: – \: 2) \: d’} [/tex]

   Taking 2 common on left side,

   • [tex] \sf \dfrac{14m \: – \: 6}{8m \: + \: 23} \: = \: \dfrac{2(a \: + \: (m \: – \: 1) \: d)}{2(a’ \: + \: (m \: – \: 1) \: d’)} [/tex]

   • [tex] \sf \dfrac{14m \: – \: 6}{8m \: + \: 23} \: = \: \dfrac{\cancel{2}(a \: + \: (m \: – \: 1) \: d)}{\cancel{2}(a’ \: + \: (m \: – \: 1) \: d’)} [/tex]

   • [tex] \sf \dfrac{14m \: – \: 6}{8m \: + \: 23} \: = \: \dfrac{a \: + \: (m \: – \: 1) \: d}{a’ \: + \: (m \: – \: 1) \: d’} [/tex]

   • [tex] \sf \dfrac{14m \: – \: 6}{8m \: + \: 23} \: = \: \dfrac{a_m}{a’_m} [/tex]

   • [tex] \boxed{\sf a_m : a’_m \: = \: 14m \: – \: 6 : 8m \: + \: 23} [/tex]

   Therefore,

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

   [tex] \huge{\bf{\purple{\mathfrak{\dag{\underline{\underline{Note:-}}}}}}} [/tex]

   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 [tex] \sf a_m [/tex] and get the ratio.

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2. Required Answer :-

   According to the question

   [tex]\sf \dfrac{Sn}{Sn’} =\dfrac{ \dfrac{n}{2}\bigg(2a + [n-1] d\bigg)}{\dfrac{n}{2}\bigg(2a’+[n-1]d’\bigg)}[/tex]

   [tex]\sf\dfrac{Sn}{Sn’} = \dfrac{\bigg(2a\bigg)[n-1]d}{\bigg(2a’\bigg)+[n-1]d’}[/tex]

   [tex]\sf\dfrac{Sn}{Sn’} = \dfrac{2a+[n-1]d}{2a’+[n-1]d’}[/tex]

   [tex]\sf \dfrac{2a+ [n – 1]d}{2a’ + [n – 1]d’}=\dfrac{7n + 1}{4n+27}[/tex]

   [tex]\sf Exchanging\; n =\; 2m-1[/tex]

   [tex]\sf \dfrac{7(2m-1)+1}{4(2m-1)+27}[/tex]

   [tex]\sf \dfrac{14m-7+1}{8m – 4 + 27}[/tex]

   [tex]\sf\dfrac{14m-6}{8m+23}[/tex]

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