# The sum of $n$ terms of two $A.P.s$ are in the ratio $5n+4:9n+6$ Find the ratio of their $18^{th}$ terms.

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• Sum of $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[2a+(n-1)d] Given that the ratio of sum of n terms of two A.P.s=\large\frac{5n+4}{9n+6} i.e., \large\frac{S_{n\:-1}}{S_{n\:-2}}=\frac{5n+4}{9n+6} Let the first terms of the A.P.s be a_1\:\:and\:\:a_2 and common differences be d_1\:\:and\:\:d_2 respectively \therefore\:\large\frac{S_{n\:-1}}{S_{n\:-2}}=\frac{\large\frac{n}{2}[2a_1+(n-1)d_1]}{\large\frac{n}{2}[2a_2+(n-1)d_2]}$$=\large\frac{5n+4}{9n+6}$
$\Rightarrow\:\large\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\large\frac{5n+4}{9n+6}$
Dividing numerator and denominator by $2$ we get
$\Rightarrow\:\large\frac{a_1+\large\frac{n-1}{2}d_1}{a_2+\large\frac{n-1}{2}d_2}=\large\frac{5n+4}{9n+6}$