Question

If the ratio of the sum of $n$ terms of two AP’s is $\left( {7n + 1} \right):\left( {4n + 27} \right)$, then the ratio of their ${11^{th}}$ term is.
$\left( 1 \right)$ $4:3$
$\left( 2 \right)$ $3:4$
$\left( 3 \right)$ $2:3$
$\left( 4 \right)$ $3:2$

Answer 1

We have to find the ratio of the ${11^{th}}$ term of the two A.P.’s then we are given the ratios of the sum of the $n$ terms of the two A.P.’s. We solve this question using the concept of the formula of the of the sum of the ${n^{th}}$ term of an A.P. and the formula of the ${n^{th}}$ term of A.P. First we will use the formula of the sum of terms of A.P. and then using the formula of the ${n^{th}}$ term of an A.P. we will simplify the expression such that we get a relation in terms of the ratios of the ${11^{th}}$ term of the two A.P.’s .

Complete step by step answer:
Given:
Let the sum of the two A.P.’s is given as $S$ and $S'$ and the ${n^{th}}$ term of the two A.P.’s are ${a_n}$ and $a{'_n}$ respectively .
Let us also consider the first term, the common difference of the two A.P. be $a$, $d$ and $a'$, $d'$ respectively.
Now, according to the consideration and the given condition, we can write the expression as:
$S':S = \left( {7n + 1} \right):\left( {4n + 27} \right)$
Also, the formula for the ${n^{th}}$ term of the A.P. is given as:
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Also, we know that the ratios can be written as:
$a:b = \dfrac{a}{b}$

Now, we can write the expression for the ratio of the sum as:
$\dfrac{{\dfrac{n}{2}\left( {2a' + \left( {n - 1} \right)d'} \right)}}{{\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)}} = \dfrac{{7n + 1}}{{4n + 27}}$
Now, cancelling the terms we can also write the expression as:
$\dfrac{{\left( {2a' + \left( {n - 1} \right)d'} \right)}}{{\left( {2a + \left( {n - 1} \right)d} \right)}} = \dfrac{{7n + 1}}{{4n + 27}}$
Now, taking $2$ common and cancelling the terms we can write the expression as:
$\dfrac{{\left( {a' + \dfrac{{\left( {n - 1} \right)}}{2}d'} \right)}}{{\left( {a + \dfrac{{\left( {n - 1} \right)}}{2}d} \right)}} = \dfrac{{7n + 1}}{{4n + 27}}$
We also know that the formula for the ${n^{th}}$ term of the A.P. is given as:
${a_n} = a + \left( {n - 1} \right)d$
As we have to find the ratio of the ${11^{th}}$ term of the A.P., we have to find the ratio of the expression given as:
$\dfrac{{a{'_{11}}}}{{{a_{11}}}} = \dfrac{{a' + \left( {11 - 1} \right)d'}}{{a + \left( {11 - 1} \right)d}}$
So on comparison of the two ratios, we conclude that we need to substitute the value as:
$\dfrac{{n - 1}}{2} = 11 - 1$
$\dfrac{{n - 1}}{2} = 10$
On solving further, we get the value as:
$n = 21$
Now on substituting the value of $n$, we can write the expression as:
$\dfrac{{\left( {a' + \dfrac{{\left( {21 - 1} \right)}}{2}d'} \right)}}{{\left( {a + \dfrac{{\left( {21 - 1} \right)}}{2}d} \right)}} = \dfrac{{7 \times 21 + 1}}{{4 \times 21 + 27}}$
$\dfrac{{\left( {a' + 10d'} \right)}}{{\left( {a + 10d} \right)}} = \dfrac{{147 + 1}}{{84 + 27}}$
Now on further solving and using the formula for the ${n^{th}}$ term of A.P. we can write the expression as:
$\dfrac{{a{'_n}}}{{{a_n}}} = \dfrac{{148}}{{111}}$
On simplifying the terms, we get the value of the ratio as:
$\dfrac{{a{'_n}}}{{{a_n}}} = \dfrac{4}{3}$

Hence, we get the ratio of the ${n^{th}}$ term of the two A.P.’s as $4:3$.

So, the correct answer is “Option 1”.

Note:
We should not take the first term and the common difference of the two A.P.’s same, as it would cancel out the left hand side completely and thus resulting in an incorrect solution. We should not get mixed up with the formula of the general terms of G.P. , A.P. and H.P.