Question

The $n^{th}$ term of a sequence is $2n-3$, find its $15^{th}$ term.

Answer 1

Hint: This question can be done by easily understanding the concepts of arithmetic progression which are mentioned below: -
$n^{th}$ term of an A.P (arithmetic progression) is given by ${T_n} = a + (n - 1)d$ where a= first term of the sequence and d=common difference which is given by $d = {T_n} - {T_{n - 1}}$. As already they have given $T_n$ of the sequence, we will just substitute the value of ‘n’ to get the required answer.

Complete step-by-step answer:
Here in this question $n^{th}$ term of a sequence is directly given i.e. (2n-3) so we can directly use this $n^{th}$ term and can find $15^{th}$ term
$ \Rightarrow {T_n} = 2n - 3$ .............equation (1)
Now we have to find the $15^{th}$ term so we will put n=15 in equation 1 to find the $15^{th}$ term.
$ \Rightarrow {T_{15}} = 2(15) - 3$
$ \Rightarrow {T_{15}} = 30 - 3$
$\therefore {T_{15}} = 27$
Therefore the $15^{th}$ term of the sequence will be 27.

Additional Information: In mathematics there are three types of progressions:-
*Arithmetic progression
*Geometric progression
*Harmonic progression
Definition of arithmetic progression: - An arithmetic sequence or progression is defined as a sequence in which for every pair of consecutive terms the second number is obtained by adding a fixed number to the first one. Difference between two consecutive terms is always a constant term.

Note: Alternate method: - We can also solve this question by using formula ${T_n} = 2n - 3$
Now we will find ${T_1}$ term by putting n=1
$ \Rightarrow {T_1} = 2(1) - 3 = - 1$
$\therefore {T_1} = - 1$
Now we will find ${T_2}$ term by putting n=2
$ \Rightarrow {T_2} = 2(2) - 3 = 1$
$\therefore {T_2} = 1$
So, common difference can be find out using formula $d = {T_n} - {T_{n - 1}}$
$ \Rightarrow d = {T_2} - {T_1}$
$ \Rightarrow d = 1 - ( - 1)$ (Putting the values)
$\therefore d = 2$
Therefore common difference is 2
Now we will find $15^{th}$ term by applying formula ${T_n} = a + (n - 1)d$
$ \Rightarrow {T_{15}} = - 1 + (15 - 1)2$ (Putting values of a=1, n=15, d=2)
$ \Rightarrow {T_{15}} = - 1 + (14)2$
$ \Rightarrow {T_{15}} = - 1 + 28$
$\therefore {T_{15}} = 27$