Question

The ratio of the sum of nth term of two AP is 7n+1/ 4n+27. Find the ratio of mth term​

Answer 1

Step-by-step explanation:

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Answer 2

$\red\bigstar$ CORRECT QUESTION:-

The sum of n terms of two aps are in the ratio (7n+1)/(4n+27). Find the ratio of mth term

$\green\bigstar$ CONCEPT:-

Here the concept of Arithematic progression has been used to find the mth term of AP .

$\blue\bigstar$ FORMULAS TO BE USED:-

Sn = n/2[2a+(n-1)d]

$\orange\bigstar$ SOLUTION:-

Let a , A be the first term and d , D be the common difference of the two given AP's .Then the sum of the nth terms are given by

$\star \sf \longrightarrow \: S_n = \dfrac{n}{2} \{2a + (n - 1)d \}$

$\star \sf \longrightarrow \: S_n^{1} = \dfrac{n}{2} \{2A+ (n - 1)D\}$

$\red{ \implies \boxed{ \frac{ \bf \: S_n}{\bf \: S_n^{1} } = \dfrac{ \bf \dfrac{n}{2} \{2a + (n - 1)d \}}{ \: \bf \dfrac{n}{2} \{2A+ (n - 1)D\}} = \bf \blue{ \dfrac{7n + 1}{4n + 27} \: \: \: |given| }}}......(1)$

Now , ratio of their mth term

$\star \sf\longrightarrow \dfrac{a + (m - 1)d}{A + (m - 1)D} = \dfrac{2a +2 (m - 1)d}{2A +2 (m - 1)D} \: \:|multiplying \: by \: 2 \:| ....(2)$

Putting  n=2m-1 in (1) , we get

$\bf \implies\dfrac{2a + (2m - 2)d}{2A + (2m - 2)D}$

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

$\purple{ \implies \bf\dfrac{14m - 6}{8m + 23}}$

$\bf \: Hence \: required \: ratio \: is \: \orange{\boxed{ \implies \bf\dfrac{14m - 6}{8m + 23}}} \bigstar$

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