Ratio of sum of n terms of two AP's is $\;(5n+3)\;:\;(3n+4)\;,$ then the ratio of $\;15^{th}\;$ term will be :

$(a)\;91:148\qquad(b)\;148:91\qquad(c)\;97:158\qquad(d)\;158:97$

Answer : (b) $148:91$
Explanation : Given , $\large\frac{\frac{n}{2}\;[2a+(n-1)d]}{\frac{n}{2}\;[2A+(n-1)D]}=\large\frac{5n+3}{3n+4}$
$\large\frac{a+\large\frac{(n-1)d}{2}}{A+\large\frac{(n-1)D}{2}}=\large\frac{5n+3}{3n+4}$
$For\;5^{th}\;term\;,$
$\large\frac{n-1}{2}=14\quad\;n=29$
$Ratio=\large\frac{5(29)+3}{3(29)+4}=\large\frac{148}{91}\;.$