[tex]\rule{200}4[/tex]
Answer : The required ratio is (14m-6):(8m+23) .
[tex]\rule{200}4[/tex]
Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .
That is.
[tex]\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\[/tex]
As , we know that the sum of n terms of an AP is given by ,
[tex]\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\[/tex]
Assume that ,
Using this we have ,
[tex]\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\[/tex]
[tex]\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\[/tex]
Now also we know that the nth term of an AP is given by ,
[tex]\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\[/tex]
Therefore,
[tex]\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\[/tex]
[tex]\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\[/tex]
From equation (i) and (iii) ,
[tex]\longrightarrow\sf\small n-1 = 2m-2\\[/tex]
[tex]\longrightarrow\sf\small n = 2m -2+1 \\[/tex]
[tex]\longrightarrow\sf\small n = 2m -1 \\[/tex]
Substitute this value in equation (i) ,
[tex]\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\[/tex]
Simplify,
[tex]\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\[/tex]
[tex]\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\[/tex]
[tex]\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\[/tex]
From equation (ii) ,
[tex]\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\[/tex]
[tex]\rule{200}4[/tex]
