Question

Which term of the AP 3, 15, 27, 39, … will be 120 more than its \[{{21}^{st}}\] term?

Answer 1

Hint: We are given an AP as 3, 15, 27, …. We find the common difference (d) and the first term (a). d is given as \[{{a}_{2}}-{{a}_{1}}\] or \[{{a}_{n}}-{{a}_{n-1}}.\] Then using them, we will find the \[{{21}^{st}}\] term using the \[{{n}^{th}}\] term formula that is, \[{{a}_{n}}=a+\left( n-1 \right)d.\] Now, we will add 120 to the \[{{21}^{st}}\] term and consider the sum as the \[{{n}^{th}}\] term and solve further for n using the formula \[{{a}_{n}}=a+\left( n-1 \right)d.\]

Complete step-by-step answer:
We are given an AP as 3, 15, 27, ……
So, its first term is a = 3 and the common difference is
\[d={{a}_{2}}-{{a}_{1}}\]
\[\Rightarrow d=15-3\]
\[\Rightarrow d=12\]
Now, we have to find a term which is 120 more than the \[{{21}^{st}}\] term.
So, first of all, we will find the \[{{21}^{st}}\] term of the AP that is given to us. We know that the \[{{n}^{th}}\] term of the AP is given as
\[{{a}_{n}}=a+\left( n-1 \right)d\]
So using this, we will get the \[{{21}^{st}}\] term as,
\[{{a}_{21}}=a+\left( 21-1 \right)d\]
\[\Rightarrow {{a}_{21}}=a+20d\]
Here, a = 3 and d = 12. So substituting these values in the above equation, we get,
\[\Rightarrow {{a}_{21}}=3+20\times 12\]
\[\Rightarrow {{a}_{21}}=3+240\]
After solving, we get,
\[\Rightarrow {{a}_{21}}=243\]
Now, we will add 120 to the \[{{21}^{st}}\] term. So, we add 120 to 243, we get,
\[243+120=363\]
Now, we will consider that our \[{{n}^{th}}\] term is 363 and we will find n. The \[{{n}^{th}}\] term is given as
\[{{a}_{n}}=a+\left( n-1 \right)d\]
So we have \[{{a}_{n}}\] as 360, a = 3 and d = 12. So, we get,
\[\Rightarrow 363=3+\left( n-1 \right)12\]
Solving further, we get,
\[\Rightarrow 363-3=\left( n-1 \right)12\]
\[\Rightarrow 360=\left( n-1 \right)12\]
Dividing both sides by 12, we get,
\[\Rightarrow \dfrac{360}{12}=\dfrac{\left( n-1 \right)12}{12}\]
\[\Rightarrow 30=n-1\]
\[\Rightarrow n=31\]
Therefore 31 is the required answer. This means that 120 more than \[{{21}^{st}}\] term is \[{{31}^{st}}\] term.

Note: While finding the \[{{n}^{th}}\] term of the AP, always remember that \[{{a}_{n}}=a+\left( n-1 \right)d.\] And for \[{{21}^{st}}\] term, we will get,
\[{{a}_{21}}=a+\left( 21-1 \right)d\]
\[\Rightarrow {{a}_{21}}=a+20d\]
Common mistakes can occur if we write \[{{a}_{21}}\] as a + 21d which is wrong. Also, while solving for n using the \[{{n}^{th}}\] term, we open the bracket of (n – 1) at the last step for easy calculation.