1. S6 = 6/2(a+5d)
d = (A7-A3)/4
a = A3-2d
2. since |r| = 1/2 < 1, yes the sum is finits
I will assume that by a3 you mean term(3)
and the terms form an geometric sequence
term(3) = ar^2 = 4/9
term(7) = ar^6 = 9/64
divide term7 by term3
r^4 = (9/64) / (4/9) = 81/256
r = ± 3/4
the in ar^2 = 4/9
a(9/16) = 4/9
a = 64/81
so Sum(6) = a(r^6 - 1)/(r-1)
= (64/81)((3/4)^6 - 1)/(3/4 - 1)
= appr 2.59799
2. term2/term1 = 1/2
term3/term2 = 1/2
since |r| < 1 , the infinite series has a finite sum
r = -7/-49 = 1/7
yes, so does that one
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