Question

in an arithmetic sequence whose first term is 4, the 1st, 3rd and 7th terms form consecutive terms of geometric sequence, find the sum of the first three terms of the arithmetic sequence

Answer 1

a + a+d + a+2d = 3 a + 3 d = 3(a+d)=12+3d

firat = 4
third = 4+2d
7th = 4+6d
r
and the geometric is
b, br, br^2, br^3 ...
4 = b
4+2d = b r = 4r
4+6d = b r^2 = 4 r^2

so
2 d = 4 r - 4 or d = 2r-2
6 d = 4 r^2 -4 or d = (2/3)r^2 -2/3

2 r - 2 = (1/3) (r^2 - 1)

6 r - 6 = r^2 - 1

r^2 - 6 r + 5 = 0
(r-5)(r-1) = 0
r = 5 or 1
1 is boring, constant terms in geo sequence
look at r = 5
d = 2 r - 2 = 8
then finally
12+3d = 12 + 24 = 36

Answer 2

picking it up from Damon's

first = 4
third = 4+2d
7th = 4+6d


but these also form a GP

(4+2d)/4 = (4+6d)/(4+2d)
16 + 16d + 4d^2 = 16 + 12d^2
8d^2 -16d = 0
8d(d - 2) = 0
d = 0 , or d = 2

for a=4, and d=2, the sum of the first three AP terms are
4 + 6 + 8
= 18

check:
term1 = 4
term3 = a+2d = 8
term7 = a+6d = 16
and 4,8, and 16 form a GP

Answer 3

In AP,a=4, 3rd=4+2d, 7th=4+6d. We are told that: 4=a in GP, 4+2d=4r in GP and 4+6d=4r^2. So it becomes: 4+2d=4r(¡) and 4+6d=4r^2(¡¡). So in (¡) d=2r-2, and substitute in (¡¡) and becomes: 4+6(2r-2)=4r^2, and becomes 4r^2-12r+8=0. Which is (r-2)(r-1)=0. So r =1 or 2 . Substituting in AP 1st term is 4 and 3rd term is 8, then suntract 4 from 8 and diide by 2 to find commom difference,which is 2. So AP terms are:4,6.8,...... So AP sum 1st 3terms= 4+6+8 =18. Sum using formula=n/2{2a+(n-1)d}, substituting it becomes 3/2(8+4)=18.

Answer 4

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