a + a r + a r^2 + a r^3 + a r^4 + a r^5 + ar^6
a r^2 = 108
a r^5 =-32
r^5/r^2 = - 32/108 = -8/27
r^3 = -2^3/3^3 =
r = - 2/3
go back and get "a" now
then if you still need help summing go here:
https://www.mathsisfun.com/algebra/sequences-sums-geometric.html
The 3rd and 6th term of a G.P are 108 and -32 respectively.Find the sum of the first 7th term.
4 answers
S7 = a(1-r^7)/(1-r)
Agree
Third term = T3
Sixth term = T6
Tn = ar^n-1
T3=ar^2=108
ar^2=108_ _ _ _(1)
T6=ar^5=-32
ar^5=-32_ _ _ _(2)
divide eqn (2) by (1)
ar^5/ar^2=-32/108
r^3=-8/27
Cuberoot b.s
r=-2/3
Substitute r into eqn (1)
a(-2/3)^2=108
a×4/9=108
a=108×9/4=27×9=243
Sum of first 7 terms = S7
S7=a(1-r^n)/1-r
=243(1-(-2/3)/1-(-2/3)
=243(1+2/3)/1+2/3
=243(5/3)/5/3
=81×5×3/5=243
Sum of first 7 terms=243
Sixth term = T6
Tn = ar^n-1
T3=ar^2=108
ar^2=108_ _ _ _(1)
T6=ar^5=-32
ar^5=-32_ _ _ _(2)
divide eqn (2) by (1)
ar^5/ar^2=-32/108
r^3=-8/27
Cuberoot b.s
r=-2/3
Substitute r into eqn (1)
a(-2/3)^2=108
a×4/9=108
a=108×9/4=27×9=243
Sum of first 7 terms = S7
S7=a(1-r^n)/1-r
=243(1-(-2/3)/1-(-2/3)
=243(1+2/3)/1+2/3
=243(5/3)/5/3
=81×5×3/5=243
Sum of first 7 terms=243