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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 !

 Oct 14, 2016
 #1
avatar+26396 
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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=82d

 

II.

aiai+2ai+4=224ai(ai+2d)(ai+4d)=224|ai=82d(82d)(82d+2d)(82d+4d)=224(82d)8(8+2d)=224|:8(82d)(8+2d)=2248(82d)(8+2d)=28|(uv)(u+v)=u2v282(2d)2=28644d2=284d2=64284d2=36|:4d2=9d=9d=3

 

III.

ai=82d|d=3ai=823ai=86ai=2

 

five consecutive terms of an arithmetic sequence: ,2,5,8,11,14,

sum: 2+5+8+11+14=40product: 2814=224

 

laugh

 Oct 14, 2016
 #1
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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

Guest Oct 14, 2016
 #3
avatar+130466 
0

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

 

 

 

cool cool cool

 Oct 14, 2016
 #4
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Thank you all !!
you were really helpfull 

 Oct 15, 2016

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