The given arithmetic progression (A.P.) is \(3, 8, 13, …, 253\).
The common difference for this A.P. is \(5\).
When this A.P. is written in reverse order, it becomes:
\(253, 248, 243, ….., 13, 8, 5\)
For this reversed A.P.:
\(a = 253\)
\(d = 248 − 253 = −5\)
\(n = 20\)
The 20th term, \(a_{20}\), is calculated as: \(a_{20 }= a + (20 − 1) d\)
Substituting the values: \(a_{20 }= 253 + (19) (−5)\)
Further calculation: \(a_{20} = 253 − 95\)
\(a_{20} = 158\)
Thus, the \(20^{th}\) term from the last term of the original A.P. is \(158\).