Question

Find out the next three terms of the given A.P:
\[\left( a+b \right),\left( a+1 \right)+b,\left( a+1 \right)+\left( b+1 \right)\].

Answer 1

Hint: The main part of an A.P is the first term and the common difference. From the given three terms we can find the common difference. We apply the formula of terms of an A.P. to find out the next terms of the series. We can also keep the exact form in which the terms are written.

Complete step-by-step solution:
The three terms of the A.P. is \[\left( a+b \right),\left( a+1 \right)+b,\left( a+1 \right)+\left( b+1 \right)\].
We express the terms of the series as ${{t}_{n}}$ which represents $n^{th}$ term of the A.P.
We define the given terms as \[{{t}_{1}}=\left( a+b \right),{{t}_{2}}=\left( a+1 \right)+b,{{t}_{3}}=\left( a+1 \right)+\left( b+1 \right)\].
Now we try to find the common difference d.
We know that ${{t}_{n}}-{{t}_{n-1}}=d$.
So, putting $n=2$ we get $d={{t}_{2}}-{{t}_{2-1}}={{t}_{2}}-{{t}_{1}}=\left( a+1 \right)+b-\left\{ \left( a+b \right) \right\}=a+1+b-a-b=1$.
The series can be found by adding 1 to the previous number of the series.
We can also keep in mind the pattern in which the terms of the series are being created.
First, we add 1 to a to get the 2nd term, then we add 1 to b to get the 3rd term…….and so on.
Also, we can use the formula for terms of ana A.P. series.
We can express ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}+(n-1)d$.
Now we find the next three terms of the series.
We add 1 to ‘’a’’ in ${{t}_{3}}$ to get the 4th term which is ${{t}_{4}}$.
So, ${{t}_{4}}={{t}_{3}}+1=\left\{ \left( a+1 \right)+1 \right\}+\left( b+1 \right)=\left( a+2 \right)+\left( b+1 \right)$.
Then we add 1 to b in ${{t}_{4}}$ to get the 5th term which is ${{t}_{5}}$.
So, ${{t}_{5}}={{t}_{4}}+1=\left( a+2 \right)+\left\{ \left( b+1 \right)+1 \right\}=\left( a+2 \right)+\left( b+2 \right)$.
We add 1 to “a” in ${{t}_{5}}$ to get the 6th term which is ${{t}_{6}}$.
So, ${{t}_{6}}={{t}_{5}}+1=\left\{ \left( a+2 \right)+1 \right\}+\left( b+2 \right)=\left( a+3 \right)+\left( b+2 \right)$.
We also maintained the pattern in which the terms are being expressed.
So, the next three terms of the A.P. are $\left( a+2 \right)+\left( b+1 \right),\left( a+2 \right)+\left( b+2 \right),\left( a+3 \right)+\left( b+2 \right)$.

Note: We need to remember that the pattern is not necessary to maintain. It’s just for the visualisation. We didn’t use the formula ${{t}_{n}}={{t}_{1}}+(n-1)d$ which we can use when we are not following the exact form of the series. The solution of the problem will always will be same.
\[{{t}_{4}}={{t}_{1}}+(4-1)d={{t}_{1}}+3=a+b+3=\left( a+2 \right)+\left( b+1 \right)\]
\[{{t}_{5}}={{t}_{1}}+(5-1)d={{t}_{1}}+4=a+b+4=\left( a+2 \right)+\left( b+2 \right)\]
\[{{t}_{6}}={{t}_{1}}+(6-1)d={{t}_{1}}+5=a+b+5=\left( a+3 \right)+\left( b+2 \right)\]
This is also another way to get the next three terms.