## College Algebra (6th Edition)

The sum of all six numbers = $\frac{1995}{64}$
Put i = 1 to 6 in $(\frac{3}{2})^{i}$ First term (i = 1) $a_{1}$ = $(\frac{3}{2})^{1}$ = $\frac{3}{2}$ Second term (i = 2) $a_{2}$ = $(\frac{3}{2})^{2}$ = $\frac{9}{4}$ Third term (i = 3) $a_{3}$ = $(\frac{3}{2})^{3}$ = $\frac{27}{8}$ Fourth term (i = 4) $a_{4}$ = $(\frac{3}{2})^{4}$ = $\frac{81}{16}$ Fifth term (i = 5) $a_{5}$ = $(\frac{3}{2})^{5}$ = $\frac{243}{32}$ Sixth term (i = 6) $a_{6}$ = $(\frac{3}{2})^{6}$ = $\frac{729}{64}$ The sum of all six numbers = $S_{56}$ = $\frac{3}{2}$ + $\frac{9}{4}$ + $\frac{27}{8}$ + $\frac{81}{16}$ + $\frac{243}{32}$ + $\frac{729}{64}$ = $\frac{(32\times3 + 16\times9 + 8\times27 + 4\times81 +2\times243 +729)}{64}$ =$\frac{1995}{64}$