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# Let , , . . . , be an arithmetic sequence. If and , then find .

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Let , . . . ,  be an arithmetic sequence. If  and , then find .

Jan 17, 2015

#1
+94548
+10

Let , . . . ,  be an arithmetic sequence. If  and , then find .

Subtracting the second sum from the first, we have

(a2 - a1) + (a4 - a3) + (a6 - a5) + (a8 - a7) + (a10 - a9) = -2   or

d           +      d       +      d       +     d        +      d        =  -2   or

5d = -2     →   d = -2/5

Also, the sum of the arithmetic series is given by

S = (10/2)(a1 + a10) = 5(a1 + a10) = 32     divide both sides by 5

So

a1 + a10 = 32/5 →  a10 = 32/5 - a1  ( 1)

And the last term is given by

a10 = a1 + d(9)    (2)

Therefore, setting (1) = (2), we have

32/5 - a1 = a1 + d(9)     ... solve for a1

(32/5 -2a1) = d(9)

32/5 -2a1 = (-2/5)(9) = -18/5

32/5 + 18/5 = 2a1

2a1 = 50/5

a1 = 5

And a10=  32/5 - a1 =  32/5 - 5 = 1.4

Proof

a1 + a3 + a5 + a7 + a9  = 5 + 4.2 + 3.4 + 2.6 + 1.8 = 17

And since each of the other 5 terms  is (2/5) less than it's preceding term.....

[17 - 5(2/5)]  = [17 - 2] = 15  = the sum of the terms with even subscripts

Jan 18, 2015

#1
+94548
+10

Let , . . . ,  be an arithmetic sequence. If  and , then find .

Subtracting the second sum from the first, we have

(a2 - a1) + (a4 - a3) + (a6 - a5) + (a8 - a7) + (a10 - a9) = -2   or

d           +      d       +      d       +     d        +      d        =  -2   or

5d = -2     →   d = -2/5

Also, the sum of the arithmetic series is given by

S = (10/2)(a1 + a10) = 5(a1 + a10) = 32     divide both sides by 5

So

a1 + a10 = 32/5 →  a10 = 32/5 - a1  ( 1)

And the last term is given by

a10 = a1 + d(9)    (2)

Therefore, setting (1) = (2), we have

32/5 - a1 = a1 + d(9)     ... solve for a1

(32/5 -2a1) = d(9)

32/5 -2a1 = (-2/5)(9) = -18/5

32/5 + 18/5 = 2a1

2a1 = 50/5

a1 = 5

And a10=  32/5 - a1 =  32/5 - 5 = 1.4

Proof

a1 + a3 + a5 + a7 + a9  = 5 + 4.2 + 3.4 + 2.6 + 1.8 = 17

And since each of the other 5 terms  is (2/5) less than it's preceding term.....

[17 - 5(2/5)]  = [17 - 2] = 15  = the sum of the terms with even subscripts

CPhill Jan 18, 2015
#2
+95356
+5

Original sequence   a=a1    d=d

$$S_n=\frac{n}{2}[2a+(n-1)d]\qquad This is the sum of an AP$$\$

 $$\\a_1+a_3+a_5+a_7+a_9=17\\ n=5,\quad a=a_1\quad, newd=2d$$     $$\\\frac{5}{2}[2a_1+(5-1)*2d]=17\\\\ \frac{5}{2}[2a_1+8d]=17\\\\ 5[a_1+4d]=17\\\\ a_1+4d=\frac{17}{5}\qquad (1)\\\\$$ $$\\a_2+a_4+a_6+a_8+a_{10}=15\\ n=5,\quad a=a_1+d, \quad newd=2d\\$$     $$\\\frac{5}{2}[2(a_1+d)+(5-1)*2d]=15\\\\ \frac{5}{2}[2(a_1+d)+8d]=15\\\\ 5[(a_1+d)+4d]=15\\\\ 5[a_1+5d]=15\\\\ a_1+5d=3\qquad (2)\\\\$$

Solve simultaneously

(2)-(1)

d=3-17/5 = -2/5

sub into (2)

a1+5(-2/5)=3

a1-2=3

a1=5

Jan 19, 2015
#3
+20848
+5

Let , . . . ,  be an arithmetic sequence.  If

and , then find .

1. arithmetic sequence: $$\small{\text{  a_1, a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10}  }}\\ \small{\text{  a_n=a_1+(n-1)*d  and  s_n= \frac{n}{2}*[2a_1*(n-1)d]  }} \small{\text{  \Rightarrow s_{10}= 17+15 = \frac{10}{2}*[2a_1+(10-1)d]  }} \\ \small{\text{  32 = 5*(2a_1+9d)  }} \\ \small{\text{  {(1) \quad 32 = 10a_1+45d}  }}$$

2. arithmetic sequence:

$$\small{\text{  a_1,a_3,a_5,a_7,a_9 }}\\ \small{\text{  a_i=a_1+(i-1)*(2d)  and  s_i= \frac{i}{2}*[2a_1*(i-1)(2d)]  }} \small{\text{  \Rightarrow s_5= 17 = \frac{5}{2}*[2a_1+(5-1)(2d)]  }} \\ \small{\text{  2*17 = 5*(2a_1+8d)  }} \\ \small{\text{  {(2) \quad 34 = 10a_1+40d}  }}$$

$$\small{\text{ (1) - (2): \quad 32-34 = 10a_1-10a_1 + 45d-40d = 5d \qquad -2=5d \qquad \boxed{d = -\frac{2}{5}}  }}$$

$$\small{\text{ d into (2):  \quad 34 = 10a_1+40*(-\frac{2}{5}) = 10a_1 - 16 \qquad 10a_1 = 50 \qquad \boxed{a_1 =5 }  }}$$

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Jan 19, 2015