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Source: — Problem Solving |
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by ssiva » Wed Mar 07, 2007 4:12 pm
240,120,60,30, 15, 15/2, 15/4, 15/8 ( cannot have 15/16 as < 1)

Hence 15/8
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Solution

by Cybermusings » Tue Mar 27, 2007 1:51 am
Answer is 15/8

It's simple

240
120
60
30
15/2 = 7.5
7.5/2 = 15/4 = 3.75
3.75/2 = 15/8 = 1.875
1.875/2 will be less than one. Hence the smallest possible number in this sequence is 15/8
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by BTGmoderatorRO » Sun Dec 03, 2017 8:13 am
240, 120, 80, 30 ...
Let the first term be a
a = 240
The sequence is a GP of common ratio r = $$\frac{1}{2}$$

Since, $$_{Tn}$$ = $$_{^{ar^{n-1}}}$$
The fifth term $$_{_{T_5}}$$ = 240 ( $$\left(\frac{1}{2}\right)$$) $$^{5-1}$$
240 ( $$\frac{1}{16}$$ ) = $$\frac{240}{16}$$
=15

Similarly,
$$_{T_6}$$ = 240/ $$_{^{2^5}}$$ = 7.5

$$_{^{_{T_7}}}$$ = 240/ $$_{^{_{^{2^6}}}}$$ =3.75

$$_{^{_{^{_{T_8}}}}}$$ = 240/ $$_{^{_{^{_{^{2^7}}}}}}$$ =1.857

$$_{^{_{^{_{^{_{_{_{T9}}}}}}}}}$$ = 240/ $$_{^{_{^{^{2^8}}}}}$$ = 0.9375
The sequence is ;
240, 120, 60, 30, 15, 7.5, 3.75, 1.857, 0.9395, ...
The least term greater than one is 1.857 = $$\frac{15}{8}$$
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by Scott@TargetTestPrep » Thu Oct 03, 2019 11:09 am
yvonne12 wrote:240,120,60,30....
in the sequence above eah term of after the first is 1/2 of the preceding term. whats the least of the sequne that is greater than 1?

32/15

16/15
15/8
15/4
15/2

pleaase expalin
After 30, we have:

15, 15/2, 15/4, 15/8, 15/16.

Since 15/16 is not greater than 1, we see that the smallest number in the sequence greater than 1 is 15/8.

Answer: C

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