Question

In an AP , the first term is $8$ , nth term is $33$ and the sum to $n$ terms is $123$ . Find the number of terms and common differences.

Answer 1

Hint: Given that the first term is $8$ and we know that the formula to find the nth term is ${t_n} = a + \left( {n - 1} \right)d$ and the sum upto to $n$ terms is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$. Substituting the given values and the first term in the above formulae we get the value of $n$ and the common difference.

Complete step-by-step solution:
Given that the first term of an AP is $8$
That is $a = 8$
We know that the $n$ th term of an AP is given by the formula
${t_n} = a + \left( {n - 1} \right)d$ …………(1)
Where $a$ is the first term and $d$ is the common difference
We are given that the $n$ th term of the AP is $33$
Substituting this and the first term in (1) we get ,
$ 33 = 8 + \left( {n - 1} \right)d \\
 \Rightarrow 33 - 8 = \left( {n - 1} \right)d $
$25 = \left( {n - 1} \right)d$ ……….(2)
We know that the sum upto to $n$ terms of an AP is given by the formula,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ ………(3)
Where $a$ is the first term and $d$ is the common difference
We are given that the sum upto $n$ terms is $123$ .
Substituting this and the first term in (3) we get
$123 = \dfrac{n}{2}\left[ {2\left( 8 \right) + \left( {n - 1} \right)d} \right]$
$\Rightarrow 123 = \dfrac{n}{2}\left[ {16 + \left( {n - 1} \right)d} \right]$…………(4)
Substituting $25 = \left( {n - 1} \right)d$ in (4) , we get
$ 123 = \dfrac{n}{2}\left[ {16 + 25} \right] \\
 \Rightarrow 123 = \dfrac{n}{2}\left[ {41} \right] $
Multiply by $2$ on both sides
$123*2 = 2*\dfrac{n}{2}\left[ {41} \right] \\
\Rightarrow 246 = 41n $
Dividing both sides by $41$ we get the value of $n$
$\dfrac{{246}}{{41}} = \dfrac{{41n}}{{41}} \\
\Rightarrow n = 6 $
Substitute the value of $n$ in (2) to get the value of $d$
$ 25 = \left( {6 - 1} \right)d \\
 \Rightarrow 25 = 5d \\
 \Rightarrow \dfrac{{25}}{5} = d \\
 \Rightarrow d = 5 $
Therefore the value of $n$ is $6$ and the common difference is $5$

Note: Many students tend to use the formula of sum upto $n$ terms as ${S_n} = \dfrac{n}{2}\left[ {a + l} \right]$. But this formula is not appropriate in this situation as the last term of the series is not given. When the initial term is given, the following terms are related to it by repeated addition of the common difference. Thus, the explicit formula is
Term = initial term + common difference * number of steps from the initial term