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Question: Find the \(9^{\text{th}}\) term from the end of the AP \(5,9,13,........185\)...

Find the 9th9^{\text{th}} term from the end of the AP 5,9,13,........1855,9,13,........185

Explanation

Solution

Find out the value of the common difference that is dd and nn which is the number of terms in the sequence from the formula Tn=a+(n−1)d{T_n} = a + (n - 1)d and here aa is the first term and Tn{T_n} is the nthn{\text{th}} term of the sequence and we now need to find the value of Tn−9{T_{n - 9}} which is 9th9^{\text{th}} term from the end.

Complete step by step solution:
Here we are given in this question that there is the sequence of AP which means Arithmetic progression here the difference between the consecutive terms is always constant and this is called as the common difference of this progression.
For example: Let us say that we have the AP a,b,c,da,b,c,d so we can say that
b−a,c−b,d−cb - a,c - b,d - c must be equal and this difference is known as the common difference of the AP. Here 5,9,13,........1855, 9, 13,........185 are said to be in AP so we get that
⇒a=5 ⇒d=9−5=13−9=4  \Rightarrow a = 5 \\\ \Rightarrow d = 9 - 5 = 13 - 9 = 4 \\\
Here we can use the formula
Tn=a+(n−1)d{T_n} = a + (n - 1)d
And here aa is the first term and Tn{T_n} is the nthn{\text{th}} term of the sequence
Tn=a+(n−1)d{T_n} = a + (n - 1)d
⇒185=5+(n−1)4 ⇒185−5=(n−1)4 ⇒1804=n−1 ⇒n=46  \Rightarrow 185 = 5 + (n - 1)4 \\\ \Rightarrow 185 - 5 = (n - 1)4 \\\ \Rightarrow \dfrac{{180}}{4} = n - 1 \\\ \Rightarrow n = 46 \\\
So there are 4646 terms in the sequence of the AP 5,9,13,........1855,9,13,........185
So 9th9^{\text{th}} term from the end is 46−8=38th term from the start46 - 8 = 38{\text{th term from the start}}
So now we get that n=38n = 38
Tn=a+(n−1)d{T_n} = a + (n - 1)d
⇒T38=5+(38−1)4 ⇒T38=5+(37)4 ⇒T38=153  \Rightarrow {T_{38}} = 5 + (38 - 1)4 \\\ \Rightarrow {T_{38}} = 5 + (37)4 \\\ \Rightarrow {T_{38}} = 153 \\\

Note:
We can do it this way also like as follows:
Nth term from the end of the AP is given by =l−(n−1)d = l - (n - 1)d
Where l,dl,d are the last term and the common difference of the arithmetic progression given.
So here l=185,n=9,d=4l = 185,n = 9,d = 4
⇒Tnth term from last=185−(9−1)4 ⇒Tnth term from last=185−32=153  \Rightarrow {T_{{\text{nth term from last}}}} = 185 - (9 - 1)4 \\\ \Rightarrow {T_{{\text{nth term from last}}}} = 185 - 32 = 153 \\\
Hence in this way also we get the same answer. So we must remember that for such questions it is necessary that we have the formula of the sum and the nth term of the AP on our tips as it helps a lot.