# The sum of n terms of two \[AP\] are in ratio $\left( {5n + 4} \right):\left( {9n + 6} \right)$ Find the ratio of their ${18^{th}}$ term?

Last updated date: 16th Mar 2023

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**Hint:**In this problem, we need to find ratio of \[{18^{th}}\text{term of first A.p}\] to that of\[{18^{th}} \text{term of second A.p}\], first we will find the number of terms that is $n$ and then we will substitute in the given equation to get the required solution.

Formulas used:

${n^{th}}$ term, ${a_n} = a + \left( {n - 1} \right)d$

Sum of $n$ terms, ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$

$a = $ first term of $A.P$

$d = $ Common difference of $A.P$

**Complete step-by-step solution:**

We need to find, ratio of \[{18^{th}}\text{term of first A.p}\] to that of \[{18^{th}}\text{term of second A.p}\]

That is \[\dfrac{{{{18}^{th}}\text{term of first A.P}}}{{{{18}^{th}}\text{term of second A.p}}} = ?\]

Let, sum of n terms of first $A.P$$ = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ and

Sum of n terms of second $A.P$ \[ = \dfrac{n}{2}\left[ {2{a^1} + \left( {n - 1} \right){d^1}} \right]\]

Where, $a = $ first term of ${1^{st}}$ $A.P$

$d = $ Common difference of ${1^{st}}$ $A.P$

${a^1} = $ First term of ${2^{nd}}$ $A.P$

${d^1} = $ Common difference of ${2^{nd}}$ $A.P$

Given: sum of n terms of two $AP$ are in ratio $ = \dfrac{{\left( {5n + 4} \right)}}{{\left( {9n + 6} \right)}}$

That is,

Sum of n terms of first $A.P$ : Sum of n terms of second $A.P$ = $\left( {5n + 4} \right):\left( {9n + 6} \right)$

\[\dfrac{{\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\dfrac{n}{2}\left[ {2{a^1} + \left( {n - 1} \right){d^1}} \right]}} = \dfrac{{\left( {5n + 4} \right)}}{{\left( {9n + 6} \right)}}\]

\[\Rightarrow \dfrac{{n\left[ {a + \dfrac{{\left( {n - 1} \right)}}{2}d} \right]}}{{n\left[ {{a^1} + \dfrac{{\left( {n - 1} \right)}}{2}{d^1}} \right]}} = \dfrac{{\left( {5n + 4} \right)}}{{\left( {9n + 6} \right)}}\]

On cancelling $n$ , we can write above equation as

\[\dfrac{{\left[ {a + \dfrac{{\left( {n - 1} \right)}}{2}d} \right]}}{{\left[ {{a^1} + \dfrac{{\left( {n - 1} \right)}}{2}{d^1}} \right]}} = \dfrac{{\left( {5n + 4} \right)}}{{\left( {9n + 6} \right)}}\] …….$\left( 1 \right)$

As we need to need to find ratio of their ${18^{th}}$ term

That is, ${18^{th}}$ term of first $A.P$ : ${18^{th}}$ term of second $A.P$ = $a + \left( {18 - 1} \right)d:{a^1} + \left( {18 - 1} \right){d^1}$

$ = \dfrac{{a + \left( {18 - 1} \right)d}}{{{a^1} + \left( {18 - 1} \right){d^1}}}$

\[ = \dfrac{{a + 17d}}{{{a^1} + 17{d^1}}}\]

Here, \[\dfrac{{n - 1}}{2} = 17\]

$n - 1 = 17 \times 2$

$n - 1 = 34$

$\therefore n = 35$

Substituting the value of $n$ in equation $\left( 1 \right)$ we get,

\[\dfrac{{\left[ {a + \dfrac{{\left( {35 - 1} \right)}}{2}d} \right]}}{{\left[ {{a^1} + \dfrac{{\left( {35 - 1} \right)}}{2}{d^1}} \right]}} = \dfrac{{\left( {5\left( {35} \right) + 4} \right)}}{{\left( {9\left( {35} \right) + 6} \right)}}\]

On simplifying, we get

\[\dfrac{{\left[ {a + \dfrac{{\left( {34} \right)}}{2}d} \right]}}{{\left[ {{a^1} + \dfrac{{\left( {34} \right)}}{2}{d^1}} \right]}} = \dfrac{{\left( {5\left( {35} \right) + 4} \right)}}{{\left( {9\left( {35} \right) + 6} \right)}}\]

Therefore, \[\dfrac{{\left[ {a + 17d} \right]}}{{\left[ {{a^1} + 17{d^1}} \right]}} = \dfrac{{\left( {5\left( {35} \right) + 4} \right)}}{{\left( {9\left( {35} \right) + 6} \right)}}\]

\[\dfrac{{\left[ {a + 17d} \right]}}{{\left[ {{a^1} + 17{d^1}} \right]}} = \dfrac{{\left( {175 + 4} \right)}}{{\left( {315 + 6} \right)}}\]

Hence, \[\dfrac{{{{18}^{th}}\text{term of first A.P}}}{{{{18}^{th}}\text{term of second A.p}}} = \dfrac{{179}}{{321}}\]

**Therefore, the ratio of ${18^{th}}$ term of ${1^{st}}$ $A.P$ and ${18^{th}}$ term of ${2^{nd}}$ $A.P$ is $179:321$**

**Note:**A progression is said to be a special type of sequence in which it is possible to obtain a formula for ${n^{th}}$ term. Arithmetic progression is the a sequence of number in which it will be having the difference of any two consecutive number will remain constant

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