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Question: The \(n^{th}\) term of a sequence is \(2n-3\), find its \(15^{th}\) term....

The nthn^{th} term of a sequence is 2n32n-3, find its 15th15^{th} term.

Explanation

Solution

This question can be done by easily understanding the concepts of arithmetic progression which are mentioned below: -
nthn^{th} term of an A.P (arithmetic progression) is given by Tn=a+(n1)d{T_n} = a + (n - 1)d where a= first term of the sequence and d=common difference which is given by d=TnTn1d = {T_n} - {T_{n - 1}}. As already they have given TnT_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 nthn^{th} term of a sequence is directly given i.e. (2n-3) so we can directly use this nthn^{th} term and can find 15th15^{th} term
Tn=2n3\Rightarrow {T_n} = 2n - 3 .............equation (1)
Now we have to find the 15th15^{th} term so we will put n=15 in equation 1 to find the 15th15^{th} term.
T15=2(15)3\Rightarrow {T_{15}} = 2(15) - 3
T15=303\Rightarrow {T_{15}} = 30 - 3
T15=27\therefore {T_{15}} = 27
Therefore the 15th15^{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 Tn=2n3{T_n} = 2n - 3
Now we will find T1{T_1} term by putting n=1
T1=2(1)3=1\Rightarrow {T_1} = 2(1) - 3 = - 1
T1=1\therefore {T_1} = - 1
Now we will find T2{T_2} term by putting n=2
T2=2(2)3=1\Rightarrow {T_2} = 2(2) - 3 = 1
T2=1\therefore {T_2} = 1
So, common difference can be find out using formula d=TnTn1d = {T_n} - {T_{n - 1}}
d=T2T1\Rightarrow d = {T_2} - {T_1}
d=1(1)\Rightarrow d = 1 - ( - 1) (Putting the values)
d=2\therefore d = 2
Therefore common difference is 2
Now we will find 15th15^{th} term by applying formula Tn=a+(n1)d{T_n} = a + (n - 1)d
T15=1+(151)2\Rightarrow {T_{15}} = - 1 + (15 - 1)2 (Putting values of a=1, n=15, d=2)
T15=1+(14)2\Rightarrow {T_{15}} = - 1 + (14)2
T15=1+28\Rightarrow {T_{15}} = - 1 + 28
T15=27\therefore {T_{15}} = 27