Question

the 4th term is (2x + 10),5th term is (4x-4) and the 6th term is (8x + 40). Then calculate for the following
(i).the value of x
(II) the first ten terms in
(a) Geometric progression
(b) Arithmetic progression

Answer 1

do we assume it arith to start out?

Answer 2

otherwise i dont see a reason why x can t be anything it wants to be

Answer 3

a sequence need not have a defining function to be a sequence

Answer 4

now, if it IS arith; then we simply can plug these into the general formula and see what comes out of it

Answer 5

suppose the a1= (2x + 10), a2= (4x-4), a3= (8x + 40) and are part of an arith seq.

then an = a1 + d(n-1)

a2 = a1 + d(2-1)

(4x-4) = (2x + 10) + 3d

2x - 14 = 3d

(2x-14)/3 = d

Answer 6

a3= a1 + d(n-1)

(8x + 40) = (2x+10) + (2x-14)(2)/3

6x + 30 = (4x-28)/3

18x + 90 = 4x-28

14x = -118

x = -118/14 = -59/7

Answer 7

good try