## College Algebra (6th Edition)

If 93 is last term of $b_{n}$ then there are 19 terms in $b_{n}$. = 19 terms
From graph First term $b_{1}$ = 3 Second term $b_{2}$ = 8 Third term $b_{3}$ = 13 Sequence = 1, 8, 13, . . . . . Common difference (d) = 13 - 8 = 8 - 3 = 5 Given Last term ($b_{L}$)= 93 Number of terms = $\frac{Last term - first term }{d}$ + 1 = $\frac{b_{L} - b_{1}}{d}$ + 1 = $\frac{93 - 3}{5}$ + 1 = $\frac{90}{5}$ + 1 = 18 + 1 = 19 If 93 is last term of $b_{n}$ then there are 19 terms in $b_{n}$.