To determine the last term (the 20th term) and the first term of the arithmetic progression (A.P) with 20 terms, a common difference of 3, and a sum of 165, you can use the following formulas:
The formula for the sum of an arithmetic series is:
Sn=n2[2a+(n−1)d]Sn​=2n​[2a+(n−1)d]
Where:

• SnSn​ is the sum of the series.
• nn is the number of terms.
• aa is the first term.
• dd is the common difference.

In this case, you know that n=20n=20, d=3d=3, and Sn=165Sn​=165. You need to find both aa (the first term) and a20a20​ (the last term).
First, let’s find aa:
165=202[2a+(20−1)⋅3]165=220​[2a+(20−1)⋅3]
Simplify this equation:
165=10[2a+19⋅3]165=10[2a+19⋅3]
165=10[2a+57]165=10[2a+57]
Now, isolate 2a+572a+57:
165=20a+570165=20a+570
Subtract 570 from both sides:
20a=165−57020a=165−570 20a=−40520a=−405
Now, divide by 20 to find aa:
a=−40520a=20−405​ a=−20.25a=−20.25
So, the first term of the A.P is -20.25.
To find the last term (a20a20​), you can use the formula for the nnth term of an A.P:
an=a+(n−1)dan​=a+(n−1)d
In this case, n=20n=20, a=−20.25a=−20.25, and d=3d=3:
a20=−20.25+(20−1)⋅3a20​=−20.25+(20−1)⋅3 a20=−20.25+19⋅3a20​=−20.25+19⋅3 a20=−20.25+57a20​=−20.25+57 a20=36.75a20​=36.75
So, the last term of the A.P is 36.75.