Question

In the given AP, find 3 missing terms:
2, _, _, _, 26.

Answer 1

Hint: In this question, we are given an arithmetic progression having 5 terms where we have to find three of the terms. We are given the first term of the arithmetic progression and the last fifth term of the arithmetic progression. We will use the formula of finding ${{n}^{th}}$ term given by ${{a}_{n}}=a+\left( n-1 \right)d$ where a is the first term, ${{a}_{n}}$ is the ${{n}^{th}}$ term and d is the common difference of the arithmetic progression. Using n = 5, we will find the value of d. After that we will find second, third and fourth term using values of a and d.

Complete step-by-step answer:
Here we are given the arithmetic progression as 2, _, _, _, 26.
We need to find missing three terms which are second, third and fourth terms of the arithmetic progression.
As we can see, the first term of the arithmetic progression is 2, so a = 2.
Here we are given the fifth term of the arithmetic progression, so let us use it to find the common difference d of the arithmetic progression.
As we know, ${{n}^{th}}$ term of an AP is given by ${{a}_{n}}=a+\left( n-1 \right)d$.
So, let us use n = 5 in the above formula to find the value of d. Value of a is given as 2 and the fifth term is 26 so, ${{a}_{5}}=26$. Hence, we get:
\[\begin{align}
  & 26=2+\left( 5-1 \right)d \\
 & \Rightarrow 26=2+\left( 4 \right)d \\
 & \Rightarrow 26-2=4d \\
 & \Rightarrow 24=4d \\
\end{align}\]
Dividing both sides by 4 we get:
\[\begin{align}
  & \Rightarrow d=\dfrac{24}{4} \\
 & \Rightarrow d=6 \\
\end{align}\]
Hence, the common difference of the given arithmetic progression is 6.
Now we need to find the missing terms which are second, third and fourth terms.
Let us find the second term given by ${{a}_{2}}$.
Now, ${{a}_{2}}=a+\left( 2-1 \right)d$.
Putting value of a and d we get:
\[\Rightarrow {{a}_{2}}=2+\left( 1 \right)6=2+6=8\]
Hence the second term is 8.
Let us find the third term given by ${{a}_{3}}$.
Now, ${{a}_{3}}=a+\left( 3-1 \right)d$.
Putting value of a and d we get:
\[\Rightarrow {{a}_{3}}=2+\left( 2 \right)6=2+12=14\]
Hence the third term is 14.
Let us find the fourth term given as ${{a}_{4}}$.
Now, ${{a}_{4}}=a+\left( 4-1 \right)d$.
Putting values of a and d we get:
\[\Rightarrow {{a}_{4}}=2+\left( 3 \right)6=2+18=20\]
Hence the fourth term is 20.
Therefore, required terms are 8, 14 and 20.
So our arithmetic progression looks like 2, 8, 14, 20, 26.

Note: Students should be careful while finding the value of d. They should note that, value of d can be negative too. Students should not get confused between n and ${{a}_{n}}$. Here, n denotes the number at which term is to be found and ${{a}_{n}}$ denotes the term at ${{n}^{th}}$ position.