Question

The fourth term of an arithmetic sequence is 141, and the seventh term is 132. The first term is _____.

Answer 1

The formula you want to use is this:

An=A1+D(N-1)

A1 is the first term of the sequence D is a common difference and N is the term of the sequence.

141= 150 - 3(4-1)
141=141

The first term of the sequence is 150

Answer 2

a,a+d,a+2d,a+3d,a+4d,... you got a+3d=141 and a+6d=132
that means 132−141=(a+6d)−(a+3d)=3d=−9 and so d=−3
The fourth term of an arithmetic sequence is 141 tells you that a+3d=141
and the seventh term is 132 means a+6d=132
from these two pieced of information we can solve for d and then solve for a

a bit of algebra shows that a+6d−(a+3d)=3d

so we see that 3d=132−141=−9

and so since it is 7-4= 3 then -9/3 would be -3 , giving us the difference

pluggin in the difference and then your equation...
a4=141
a3=141+3=144
a2=144+3=147
a1=147+3=150
so then the first term would be 150