If the  ratio of the  sum  of n terms of two  AP's be (7n  + 1):(4n + 27),  then the  ratio of their 11th  terms will be

Let S_n, and S' be the sums of n terms of two AP's and T₁₁ and T₁₁, be the respective 11th term, then

$\frac{{\mathrm{S}}_{\mathrm{n}}}{{\mathrm{S}}_{\mathrm{n}}^{\mathrm{\prime }}}=\frac{\frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1\left)\mathrm{d}\right]}{\frac{\mathrm{n}}{2}\left[2{\mathrm{a}}^{\mathrm{\prime }}+(\mathrm{n}-1){\mathrm{d}}^{\mathrm{\prime }}\right]}=\frac{7\mathrm{n}+1}{4\mathrm{n}+27}$ (given)

$\Rightarrow$$\frac{\mathrm{a}+\frac{(\mathrm{n}-1)}{2}\mathrm{d}}{{\mathrm{a}}^{\mathrm{\prime }}+\frac{(\mathrm{n}-1)}{2}{\mathrm{d}}^{\mathrm{\prime }}}=\frac{7\mathrm{n}+1}{4\mathrm{n}+27}$ 

Now, put n=21, we get 

$\frac{\mathrm{a}+10\mathrm{d}}{{\mathrm{a}}^{\mathrm{\prime }}+10{\mathrm{d}}^{\mathrm{\prime }}}=\frac{{\mathrm{T}}_{11}}{{\mathrm{T}}_{11}^{\mathrm{\prime }}}=\frac{148}{111}=\frac{4}{3}$