The sums of n terms of two A.P.'s …

Let the Sum of n^th term of first AP be {tex}{{\rm{S}}_n}{/tex} and that of second term be S'_n.

=>     {tex}{{{n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]} \over {{n \over 2}\left[ {2a' + \left( {n - 1} \right)d'} \right]}} = {{5n + 4} \over {9n + 6}}{/tex}

=>     {tex}{{2a + \left( {n - 1} \right)d} \over {2a' + \left( {n - 1} \right)d'}} = {{5n + 4} \over {9n + 6}}{/tex}

Putting n = 35, we get

=>     {tex}{{2a + \left( {35 - 1} \right)d} \over {2a' + \left( {35 - 1} \right)d'}} = {{5 \times 35 + 4} \over {9 \times 35 + 6}}{/tex}

=>     {tex}{{2a + 34d} \over {2a' + 34d'}} = {{175 + 4} \over {315 + 6}}{/tex}

=>     {tex}{{a + 17d} \over {a' + 17d'}} = {{179} \over {321}}{/tex}

=>     {tex}{{a + \left( {18 - 1} \right)d} \over {a' + \left( {18 - 1} \right)d'}} = {{179} \over {321}}{/tex}

=>     {tex}{{{a_{18}}} \over {a{'_{18}}}} = {{179} \over {321}}{/tex}

Therefore, the ratio of 18th term of two APs is 179 : 321.