Question

In the following AP, find the missing terms:
$\_,38,\_,\_,\_,-22$

Answer 1

Hint: In this problem, at first, we write the general expression of an AP which is ${{a}_{n}}=a+\left( n-1 \right)d$ . After that, since we are given the second and the sixth terms, we put $n=2,6$ separately in the general expression and then subtract the two equations to get the value of a and d. Having found these values, we then gradually put $n=3,4,5$ in the general expression to find out the missing terms.

Complete step by step solution:
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 given problem, it is said that the ${{2}^{nd}}$ term is $38$ ; ${{6}^{th}}$ term is $-22$ . So, putting ${{a}_{n}}=38,n=2$ and then ${{a}_{n}}=-22,n=6$ in equation $\left( i \right)$ , we get,
\[38=a+\left( 2-1 \right)d....\left( ii \right)\]
$-22=a+\left( 6-1 \right)d....\left( iii \right)$
Subtracting $\left( iii \right)$ from $\left( ii \right)$ , we get,
$\begin{align}
  & \Rightarrow 60=\left( 2-6 \right)d \\
 & \Rightarrow d=-12 \\
\end{align}$
Putting this value of d in $\left( ii \right)$ , we get,
$\begin{align}
  & \Rightarrow 38=a+\left( -12 \right) \\
 & \Rightarrow a=50 \\
\end{align}$
Putting these values of a and d in $\left( i \right)$ , we get,
$\begin{align}
  & \Rightarrow {{a}_{n}}=50+\left( n-1 \right)\left( -12 \right) \\
 & \Rightarrow {{a}_{n}}=-12n+62....\left( iv \right) \\
\end{align}$
Putting $n=3,4,5$ in equation $\left( iv \right)$ , we get,
$\begin{align}
  & {{a}_{3}}=-12\left( 3 \right)+62,{{a}_{4}}=-12\left( 4 \right)+62,{{a}_{5}}=-12\left( 5 \right)+62 \\
 & \Rightarrow {{a}_{3}}=26,{{a}_{4}}=14,{{a}_{5}}=2 \\
\end{align}$
Therefore, we can conclude that the AP is $50,38,26,14,2,-10,-22$.

Note: Problems on sequence and series are quite simple lest the common difference is negative, such as in this case. So, we need to be extra careful over here. Also, this problem can be solved in another way. Since, we are given the second and the sixth terms, the common difference can be found out directly by subtracting the two terms and then dividing the difference by $5$ .