 QUESTION

# Find the ${n^{th}}$ term of the AP given by 9, 13, 17, 21, 25, ….

Hint: Here, we are given an AP, so we will find out the first term $a$, common difference $d$ to calculate the ${n^{th}}$ term of the AP. Also we know ${n^{th}}$ term of an AP is given by ${T_n} = a + (n - 1)d$ where $a$ is the frost term, $d$ is the common difference, $n$ denotes the number of terms and ${T_n}$ denotes the ${n^{th}}$ term of an AP, given as $a,a + d,a + 2d,a + 3d....$
According to question, we have $a = {T_1} = 9,{T_2} = 13,{T_3} = 17....$
$a = 9$ and $d = {T_2} - {T_1} \Rightarrow 13 - 9 = 4$
Now, we know ${T_n} = a + (n - 1)d$ for ${n^{th}}$ term
$\therefore$ ${T_n} = 9 + (n - 1)4$
$\Rightarrow {T_n} = 9 + 4n - 4 \Rightarrow 5 + 4n$
Hence, we get the ${n^{th}}$ term of the given AP, ${T_n} = 5 + 4n$.
Note: For solving a question based on the ${n^{th}}$ term of an AP, we need to solve it as explained above. Also if sum of $n$ terms is asked, then we can solve it by using the summation formula for $n$terms as given below:
${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right) \Rightarrow \dfrac{n}{2}({T_1} + {T_n})$ where ${T_1}$ and ${T_4}$are the first and last term of an AP.
Here we need to keep in mind that the ${n}^th$ term and the sum of ${n}^{th}$ term are different things.