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# Arithmetic Sequence Problem Set

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1. A certain 300-term geometric sequence has first term 1337 and common ratio -1/2. How many terms of this sequence are greater than 1?

2. Let a_1, a_2, . . . ,a_10 be an arithmetic sequence. If a_1 + a_3 + a_5 + a_7 + a_9 = 17 and a_2 + a_4 + a_6 + a_8 + a_{10} = 15, then find a_1.

3. Consider this pattern where the positive, proper fractions with denominator (n+1) are arranged in the nth row in a triangular formation. The 1st through 4th rows are shown; each row has one more entry than the previous row. What is the sum of the fractions in the 15th row?

higgsb  Sep 20, 2016

#2
+17746
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1)  First term  =  1337  and the common ratio is  -1/2:

Terms are:  1337,  -668.5,  334.25,  -167.125,  83.5625,  -41.78125,  20.890625,  -10.4453125,  5.22265625,

-2.611328125,  1.3056640625, -0.65283203125

and all the remaining terms are smaller than 1.

So, to find the answer, add the positive terms in the list above.

2)  a1 + a3 + a5 + a7 + a9  =  17

--->     a1 + (a1​ + 2d) + (a1​ + 4d) + (a1​ + 6d) + (a1​ + 8d)  =  17

--->     5a1 + 20d  =  17

a2 + a4 + a6 + a8 + a10  =  15

--->   (a1 + d) + (a1​ + 3d) + (a1​ + 5d) + (a1​ + 7d) + (a1​ + 9d)  =  15

--->     5a1 + 25d  =  15

Combining these answers:          5a1 + 20d  =  17

5a1 + 25d  =  15

Subtracting down:                                 -5d  =  2

d  =  -2/5

Since  5a1 + 20d  =  17     --->     5a1 + 20(-2/5)  =  17

--->     5a1 - 8 =  17

--->     5a1 =  25

--->        a1 =  5

geno3141  Sep 20, 2016
#1
0

Ok so for the first question it is simple because all you have to do is crunch the number which can be simplified which will equal 83 terms that are 1

Guest Sep 20, 2016
#2
+17746
+10

1)  First term  =  1337  and the common ratio is  -1/2:

Terms are:  1337,  -668.5,  334.25,  -167.125,  83.5625,  -41.78125,  20.890625,  -10.4453125,  5.22265625,

-2.611328125,  1.3056640625, -0.65283203125

and all the remaining terms are smaller than 1.

So, to find the answer, add the positive terms in the list above.

2)  a1 + a3 + a5 + a7 + a9  =  17

--->     a1 + (a1​ + 2d) + (a1​ + 4d) + (a1​ + 6d) + (a1​ + 8d)  =  17

--->     5a1 + 20d  =  17

a2 + a4 + a6 + a8 + a10  =  15

--->   (a1 + d) + (a1​ + 3d) + (a1​ + 5d) + (a1​ + 7d) + (a1​ + 9d)  =  15

--->     5a1 + 25d  =  15

Combining these answers:          5a1 + 20d  =  17

5a1 + 25d  =  15

Subtracting down:                                 -5d  =  2

d  =  -2/5

Since  5a1 + 20d  =  17     --->     5a1 + 20(-2/5)  =  17

--->     5a1 - 8 =  17

--->     5a1 =  25

--->        a1 =  5

geno3141  Sep 20, 2016
#3
0

If you keep on mltiplying by -.5, you will eventually get to 6 POSITIVE ones that actually count.

Guest Feb 20, 2017