If the ratio of the sum …

Let a₁ and a₂ be the first terms and d₁ and d₂ be the common difference of the two APs respectively.
Let S_n and S'_n be the sums of the first n terms of the two APs and T_n and T'_n be their nth terms respectively.
Then, {tex}\frac { S _ { n } } { S _ { n } ^ { \prime } } = \frac { 7 n + 1 } { 4 n + 27 } \Rightarrow \frac { \frac { n } { 2 } \left[ 2 a _ { 1 } + ( n - 1 ) d _ { 1 } \right] } { \frac { n } { 2 } \left[ 2 a _ { 2 } + ( n - 1 ) d _ { 2 } \right] } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
{tex}\Rightarrow \frac { 2 a _ { 1 } + ( n - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( n - 1 ) d _ { 2 } } = \frac { 7 n + 1 } { 4 n + 27 }{/tex} ........(i)
To find the ratio of mth terms, we replace n by (2m -1) in the above expression.
Replacing n by (2 {tex}\times{/tex} 9 -1), i.e., 17 on both sides in (i), we get
{tex}\frac { 2 a _ { 1 } + ( 17 - 1 ) d _ { 1 } } { 2 a _ { 2 } + ( 17 - 1 ) d _ { 2 } } = \frac { 7 \times 17 + 1 } { 4 \times 17 + 27 } \Rightarrow \frac { 2 a _ { 1 } + 16 d _ { 1 } } { 2 a _ { 2 } + 16 d _ { 2 } } = \frac { 120 } { 95 }{/tex}
{tex}\Rightarrow \frac { a _ { 1 } + 8 d _ { 1 } } { a _ { 2 } + 8 d _ { 2 } } = \frac { 24 } { 19 }{/tex}
{tex}\Rightarrow \frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \frac { 24 } { 19 }{/tex}
{tex}\Rightarrow \frac { a _ { 1 } + ( 9 - 1 ) d _ { 1 } } { a _ { 2 } + ( 9 - 1 ) d _ { 2 } } = \frac { 24 } { 19 }{/tex}
{tex}\Rightarrow \frac { T _ { 9 } } { T _ { 9 } ^ { \prime } } = \frac { 24 } { 19 }{/tex}
{tex}\therefore{/tex} required ratio = 24:19.