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Questions  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Q)

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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A)
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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
$\Rightarrow\:\large\frac{a_1+\large\frac{n-1}{2}d_1}{a_2+\large\frac{n-1}{2}d_2}=\large\frac{5n+4}{9n+6}$
By putting $\large\frac{n-1}{2}$$=17$ we get
$n=35$
$\Rightarrow\:\large\frac{a_1+17d_1}{a_2+17d_2}=\large\frac{5n+4}{9n+6}=\frac{5\times 35+4}{9\times 35+6}$
We know that in an A.P. $t_n=a+(n-1)d$
$t_{18}=a+17d$
$\therefore\:\large\frac{t_{18_1}}{t_{18_2}}=\frac{179}{321}$
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