Courses
Courses for Kids
Free study material
Free LIVE classes
More # 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
Total views: 203.4k
Views today: 3.83k Verified
203.4k+ views
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