Question

Find the 3rd term of an arithmetic sequence with t5 = 3 and t7 = 7.

Answer 1

tn=t1+d(n-1)
t5-t7=2d
7-3=4
2d=4
d=2

tn=t1+2(n-1)
t5=3
3=t1+2(5-1)
3=t1+2(4)
3=t1+8
minus 8
-5=t1

tn=-5+2(n-1)
t3=-5+2(3-1)
t3=-5+2(2)
t3=-5+4
t3=-1

3rd term is -1