The sums of $n$ terms of two arithmetic progressions are in the ratio $5 n+4: 9 n+6 .$ Find the ratio of their $18^{\text {th }}$ terms.
The sums of $n$ terms of two arithmetic progressions are in the ratio $5 n+4: 9 n+6 .$ Find the ratio of their $18^{\text {th }}$ terms.
Let $a_{1}, a_{2}$ and $d_{1}, d_{2}$ be the first terms and the common difference of the first and second arithmetic progression respectively.
According to the given condition,
$\frac{{{\rm{ Sum }}\,\,{\rm{of }}\,\,n\,\,{\rm{ terms }}\,\,{\rm{of}}\,\,{\rm{ first}}\,\,{\rm{ A}}{\rm{.P}}{\rm{. }}}}{{{\rm{ Sum}}\,\,{\rm{ of }}\,\,n{\rm{ }}\,\,{\rm{terms }}\,\,{\rm{of }}\,\,{\rm{second}}\,\,{\rm{ A}}{\rm{.P}}{\rm{. }}}} = \frac{{5n + 4}}{{9n + 6}}$
$\Rightarrow \frac{\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]}{\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]}=\frac{5 n+4}{9 n+6}$
$\Rightarrow \frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{5 n+4}{9 n+5}$ ..........$(1)$
Substituting $n=35$ in $(1),$ we obtain
$\frac{2 a_{1}+34 d_{1}}{2 a_{2}+34 d_{2}}=\frac{5(35)+4}{9(35)+6}$
$\Rightarrow \frac{a_{1}+17 d_{1}}{a_{2}+17 d_{2}}=\frac{179}{321}$ ...........$(2)$
$\frac{{{{18}^{th}}\,\,{\rm{ term}}\,\,{\rm{of}}\,\,{\rm{ first }}}}{{{{18}^{th}}\,\,{\rm{ term }}\,\,{\rm{of }}\,\,{\rm{second}}\,\,{\rm{ A}}{\rm{.P}}{\rm{. }}}} = \frac{{{a_1} + 17{d_1}}}{{{a_2} + 17{d_2}}}$ ............$(3)$
From $(2)$ and $(3),$ we obtain
$\frac{{{{18}^{{\rm{th }}}}\,\,{\rm{ term}}\,\,{\rm{ of }}\,\,{\rm{first }}}}{{{{18}^{{\rm{th }}}}\,\,{\rm{ term }}\,\,{\rm{of}}\,\,{\rm{ second }}\,\,{\rm{A}}{\rm{.P}}{\rm{. }}}} = \frac{{179}}{{321}}$
Thus, the ratio of $18^{\text {th }}$ term of both the $A.P.$s is $179: 321 .$
Similar Questions
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