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?
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
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
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