Question

The first and last term of an A.P. of $n$ terms is $1$ , $\;31$ respectively. The ratio of ${8^{th}}$ term and ${(n - 2)^{th}}$ term is $\;5:9$ , the value of $n$ .

Answer 1

Hint: Here, we have given an A.P. of $n$ terms and also the first and last term of the A.P. We can substitute the last term equal to $\;31$ and find the common difference. Then, substitute it in the ratio of the terms and hence we can find the value of n.

Formula used: ${n^{th}}$ term of an A.P. is ${a_n} = a + (n - 1)d$ where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms in an A.P.

Complete step-by-step solution:
Let ${a_1},{a_2},{a_3},...,{a_n}$ be an A.P.
It is given that ${a_1} = 1$ and ${a_n} = 31$ . Therefore, the A.P. can be written as
$1,{a_2},{a_3},...,\;{a_{n - 1}},31$
Now, from the above A.P., we have
${a_1} = 1$ , ${a_n} = 31$ and $n$ is the number of terms
We take, ${a_n} = 31$ - - - - - - - - - - - $(1.)$
We know that,
${a_n} = a + (n - 1)d$
We substitute ${a_1} = 1$ and ${a_n} = 31$ in $(1.)$ ,
$\Rightarrow 31 = 1 + (n - 1)d$
$\Rightarrow 31 - 1 = (n - 1)d$
Simplifying the left-hand side, we get
$\Rightarrow 30 = (n - 1)d$
$\Rightarrow d = \dfrac{{30}}{{n - 1}}$ - - - - - - - - - - - - $(2.)$
Which is the common difference.
Now, according to the question, we have
$\dfrac{{{a_8}}}{{{a_{n - 2}}}} = \dfrac{5}{9}$
By using the formula of the last term of an A.P., we can find the required terms
$\Rightarrow \dfrac{{a + 7d}}{{a + (n - 3)d}} = \dfrac{5}{9}$
Now, by substituting ${a_1} = 1$, we get
$\Rightarrow \dfrac{{1 + 7d}}{{1 + (n - 3)d}} = \dfrac{5}{9}$
Cross multiplying the above fractions, we get
$\Rightarrow 9(1 + 7d) = 5(1 + (n - 3)d)$
$\Rightarrow 9 + 63d = 5 + 5(n - 3)d$
Taking terms containing $d$ on the left-hand side,
$\Rightarrow 63d - 5(n - 3)d = 5 - 9$
$\Rightarrow d(63 - 5(n - 3)) = - 4$
Taking $d$ common and simplifying the terms inside the brackets,
$\Rightarrow d(63 - 5n + 15) = - 4$
$\Rightarrow d(78 - 5n) = - 4$
Here, by substituting $d$using $(2.)$
$\Rightarrow \left( {\dfrac{{30}}{{n - 1}}} \right)(78 - 5n) = - 4$
By multiplying $n - 1$ on the right-hand side, we get
$\Rightarrow 30(78 - 5n) = - 4(n - 1)$
$\Rightarrow 2340 - 150n = - 4n + 4$
Taking terms containing $n$ on the left-hand side and other terms on the right-hand side,
$\Rightarrow - 150n + 4n = 4 - 2340$
$\Rightarrow - 146n = - 2336$
Simplifying further and evaluating the value of $n$ , we get
$\Rightarrow n = \dfrac{{ - 2336}}{{ - 146}}$
$\Rightarrow n = 16$

Therefore the value of n is equal to 16.

Note: Here, $n$ represents the total number of terms that an A.P. contains, whereas ${a_n}$ represents the last term or ${n^{th}}$ term of an A.P. Using the ${a_n}$ term, we can find any term required of the A.P. We substitute the required number of the term in place of $n$ and evaluate the required term.