Answer :

[tex]\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ n=76\\ d=4.8\\ a_{76}=375 \end{cases} \\\\\\ a_{76}=a_1+(76-1)(4.8)\implies 375=a_1+(76-1)(4.8) \\\\\\ 375=a_1+(75)(4.8)\implies 375=a_1+360\implies 15=a_1[/tex]
If the 76th term is 375, one must subtract the common difference n-1 times to get the first term. n is 76 so 375-(75*4.8) which is 15 and we plug 15 into the equation to see if we are right. A(76)=15+(76-1)*4.8 and since we get 375 we are correct. 

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