Question

How to find the first term, the common difference, and the ${{n}^{th}}$ term of the arithmetic sequence described below? $\left( 1 \right)$ ${{4}^{th}}$ term is $11$ ; ${{10}^{th}}$ term is $29$ $\left( 2 \right)$ ${{8}^{th}}$ term is $4$ ; ${{18}^{th}}$ term is $-96$ ?

Answer 1

At first, we write the general expression for the ${{n}^{th}}$ term of an arithmetic sequence. After that, we simply put the values of ${{a}_{n}}$ and its corresponding n in this general expression and then subtracting the two equations, we get the value of d, the common difference. Putting the value of d in one of the two equations, we get the value of a, the first term and finally the ${{n}^{th}}$ .

Complete step by step answer:
A sequence is an enumerated collection of objects, especially numbers, in which repetitions are allowed and in which the order of objects matters. A sequence may be finite or infinite depending on the number of objects in the sequence. Sequences can be of various types such as arithmetic sequence, geometric sequence and so on. Sequences can be completely random as well. The ${{n}^{th}}$ term of a sequence is sometimes written as a function of n.
An arithmetic sequence is the one in which the arithmetic difference between the consecutive numbers is constant. The expression for the ${{n}^{th}}$ term of an arithmetic sequence is,
${{a}_{n}}=a+\left( n-1 \right)d....\left( i \right)$
Where a is the first term of the sequence and d is the common difference.
For the first part, it is said that the ${{4}^{th}}$ term is $11$ ; ${{10}^{th}}$ term is $29$ . So, putting ${{a}_{n}}=11,n=4$ and then ${{a}_{n}}=29,n=10$ in equation $\left( i \right)$ , we get,
$11=a+\left( 4-1 \right)d....\left( ii \right)$
$29=a+\left( 10-1 \right)d....\left( iii \right)$
Subtracting $\left( ii \right)$ from $\left( iii \right)$ , we get,
$\begin{aligned} & \Rightarrow 18=\left( 9-3 \right)d \\\ & \Rightarrow d=3 \\\ \end{aligned}$
Putting this value of d in $\left( ii \right)$ , we get,
$\begin{aligned} & \Rightarrow 11=a+\left( 3 \right)3 \\\ & \Rightarrow a=2 \\\ \end{aligned}$
Putting these values of a and d in $\left( i \right)$ , we get,
$\begin{aligned} & \Rightarrow {{a}_{n}}=2+\left( n-1 \right)3 \\\ & \Rightarrow {{a}_{n}}=3n-1 \\\ \end{aligned}$
Therefore, we can conclude that the first term is $2$ , the common difference is $3$ and the ${{n}^{th}}$ term is ${{a}_{n}}=3n-1$ .
For the second part, it is said that the ${{8}^{th}}$ term is $4$ ; ${{18}^{th}}$ term is $-96$ . So, putting ${{a}_{n}}=4,n=8$ and then ${{a}_{n}}=-96,n=18$ in equation $\left( i \right)$ , we get,
$4=a+\left( 8-1 \right)d....\left( iv \right)$
$-96=a+\left( 18-1 \right)d....\left( v \right)$
Subtracting $\left( iv \right)$ from $\left( v \right)$ , we get,
$\begin{aligned} & \Rightarrow -100=\left( 17-7 \right)d \\\ & \Rightarrow d=-10 \\\ \end{aligned}$
Putting this value of d in $\left( iv \right)$ , we get,
$\begin{aligned} & \Rightarrow 4=a+\left( 8-1 \right)\left( -10 \right) \\\ & \Rightarrow a=74 \\\ \end{aligned}$
Putting these values of a and d in $\left( i \right)$ , we get,
$\begin{aligned} & \Rightarrow {{a}_{n}}=74+\left( n-1 \right)\left( -10 \right) \\\ & \Rightarrow {{a}_{n}}=-10n+84 \\\ \end{aligned}$
Therefore, we can conclude that the first term is $74$ , the common difference is $-10$ and the ${{n}^{th}}$ term is ${{a}_{n}}=-10n+84$ .

Note: In these types of problems, we first be careful while writing the expression for the ${{n}^{th}}$ term. This problem can also be solved by a sort of shortcut. Suppose, if we are given the ${{5}^{th}}$ term and the ${{9}^{th}}$ term, then the common difference will be $\dfrac{\left( {{9}^{th}}~term \right)-\left( {{5}^{th}}~term \right)}{9-5}$ .