Answer:
Find the 40th term for the arithmetic sequence in which
a8=60 and a12=48 .
Substitute 60 for a8 and 48 for a12 in the formula
an=a1+(n−1)d to obtain a system of linear equations in terms of a1 and d .
a8=a1+(8−1)d→60=a1+7da12=a1+(12−1)d→48=a1+11d
Subtract the second equation from the first equation and solve for d .
12=−4d−3=d
Then 60=a1+7(−3) . Solve for a .
60=a1−2181=a1
Now use the formula to find a40 .
a40=81+39(−3)=81−117=−36 .
Step-by-step explanation:
