A=T1=3+(2n/3)=3+{(2×1)/3}=11/3l=T31=3+(2n/3)=3+{(2×31)/3}=71/3Now, S31=(n/2) (a+l)=(31/2) {(11/3)+(71/3)}=(31×41)/3=1271/3=423.67

Sum of the First 31 Terms of an Arithmetic Progression
To find the sum of the first 31 terms of an arithmetic progression (AP) whose nth term is given by 3 + 2n/3, we need to first determine the common difference of the AP.

Finding the Common Difference
The nth term of an AP is given by the formula: \(a_n = a_1 + (n-1)d\), where \(a_n\) is the nth term, \(a_1\) is the first term, n is the term number, and d is the common difference.
Given that the nth term is 3 + 2n/3, we can equate this to the general formula to find the common difference:
\(3 + \frac{2n}{3} = a_1 + (n-1)d\)
Comparing coefficients, we get:
\(a_1 = 3\) and \(d = \frac{2}{3}\)
Therefore, the common difference of the AP is 2/3.

Finding the Sum of the First 31 Terms
The sum of the first n terms of an AP can be calculated using the formula: \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\)
Substitute the values of \(a_1\), \(d\), and \(n = 31\) into the formula to find the sum of the first 31 terms:
\(S_{31} = \frac{31}{2}[2(3) + (31-1)\left(\frac{2}{3}\right)]\)
\(S_{31} = \frac{31}{2}[6 + 30\left(\frac{2}{3}\right)]\)
\(S_{31} = \frac{31}{2}[6 + 20]\)
\(S_{31} = \frac{31}{2}(26)\)
\(S_{31} = 403\)
Therefore, the sum of the first 31 terms of the given AP is 403.