Given :
[tex]\begin{gathered} a_1=_{}10\text{ } \\ S_{30}\text{ = }4350 \end{gathered}[/tex]
Calculation :
The sum of the arithmetic progression is given as,
[tex]\begin{gathered} S_n=\text{ }\frac{n}{2}\lbrack2a_1\text{ + ( n - 1 )d \rbrack} \\ S_{30}\text{ =}\frac{30}{2}\text{ \lbrack 2}\times\text{ 10 + ( 30 - 1 ) d \rbrack} \\ \end{gathered}[/tex]
Further,
[tex]\begin{gathered} 4350\text{ = 15 }\times\text{ \lbrack 20 + 29d \rbrack} \\ 20\text{ + 29 d = }\frac{4350}{15} \\ 20\text{ + 29d = }290 \\ 29d\text{ = 290 - 20} \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} d\text{ = }\frac{270}{29} \\ d\text{ = 9.31} \end{gathered}[/tex]
The 30th term of the given sequence is calculated as,
[tex]\begin{gathered} a_{30}=a_1\text{ + ( 30 - 1 )9.31} \\ a_{30\text{ }}=\text{ 10 + 29 }\times\text{ 9.31} \\ a_{30\text{ }}=\text{ 10 + 269.99} \\ a_{30\text{ }}=\text{ 280} \end{gathered}[/tex]
Thus the 30th term of the given sequence is 280.
