five consecutive terms of an arithmetic sequence have a sum of 40. The product of the first, middle and the last term is 224. Find the terms of sequence PLEASE HELP !
five consecutive terms of an arithmetic sequence have a sum of 40. The product of the first, middle and the last term is 224. Find the terms of sequence PLEASE HELP !
I.
ai+ai+1+ai+2+ai+3+ai+4=40ai+(ai+d)+(ai+2d)+(ai+3d)+(ai+4d)=405ai+10d=40|:5ai+2d=8ai=8−2d
II.
ai⋅ai+2⋅ai+4=224ai⋅(ai+2d)⋅(ai+4d)=224|ai=8−2d(8−2d)⋅(8−2d+2d)⋅(8−2d+4d)=224(8−2d)⋅8⋅(8+2d)=224|:8(8−2d)⋅(8+2d)=2248(8−2d)⋅(8+2d)=28|(u−v)⋅(u+v)=u2−v282−(2d)2=2864−4d2=284d2=64−284d2=36|:4d2=9d=√9d=3
III.
ai=8−2d|d=3ai=8−2⋅3ai=8−6ai=2
five consecutive terms of an arithmetic sequence: …,2,5,8,11,14,…
sum: 2+5+8+11+14=40product: 2⋅8⋅14=224
The Series is: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32......etc. The common difference =3
2 x 8 x 14=224
2+5+8+11+14=40
Let the terms be a, a + d, a + 2d, a + 3d, a + 4d = 40....simplify
5a + 10d = 40 divide through by 5
a + 2d = 8 → a = 8 - 2d
And we know that
a (a + 2d)(a + 4d) = 224 substitute for a
(8 - 2d)(8 - 2d + 2d)(a - 2d + 4d) = 224 simplify
(8 - 2d)(8)(8 + 2d) = 224 divide through by 8
(8 - 2d)(8 + 2d) = 28 simplify
64 - 4d^2 = 28
4d^2 - 36 = 0
d^2 - 9 = 0 factor
(d + 3) (d - 3) = 0 do d = 3 or d = -3 reject - 3
So.....a = 8 - 2d → 8 - 2(3) = 2
And the terms are
2 + 5 + 8 + 11 + 14 = 40