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Sequence

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Re: Sequence

by parallel_chase » Mon Oct 27, 2008 8:46 am
smohsin wrote:In a sequence t1, t2, t3, t4, t5......tn, t1=23. If Tn=T(n-1) -3 for each n>1 what is tn=-4?

Please help in solving.
T2 = T1-3 = 23-3 = 20

Similarly,

Tn = 23 - 3x = -4

23+4 = 3x

x = 27/3 = 9

therefore, Tn = x+1 = 9+1 = 10


Hope this helps.
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by smohsin » Mon Oct 27, 2008 9:01 am
Dear parallel-chase,

could you please elaborate how you got :

Tn = 23 - 3x = -4
and

why you added x+1, please
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by parallel_chase » Mon Oct 27, 2008 9:21 am
smohsin wrote:Dear parallel-chase,

could you please elaborate how you got :

Tn = 23 - 3x = -4
and

why you added x+1, please
T1 = 23

Tn = T(n-1) - 3

T2 = T1 - 3 = 23-3 = 20
T3 = T2 - 3 = 20-3 = 17
T4 = T3 - 3 = 17-3 = 14
T5 = T4 - 3 = 14-3 = 11
T6 = T5 - 3 = 11-3 = 8
T7 = T6 - 3 = 8-3 = 5
T8 = T7 - 3 = 5-3 = 2
T9 = T8 - 3 = 2-3 = -1
T10 = T9 -3 = -1-3 = -4


As you can see from the above pattern that we are subtracting 3 from every number.

Hence easy way of doing this is

Tn = 23 - 3x = -4

Here x is the number of times 3 has been subtracted.

We added 1 to the value of x because of inclusiveness. Because when calculating the sequence we do not include the first value.

Hope this helps.
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by vishubn » Mon Oct 27, 2008 9:23 am
alternate approach

get the series first ....

T1=23

so with the given condition

we get T2=T1-3=20
T3=17 and so on

the series is 23,20,17,.....
now using the formula

tn=a+(n-1)d

where tn=-4,a=23 and d=-3

-4=23+(n-1)*-3
-4=23-3n+3
3n=30
n=10


HTh
Vishu
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by smohsin » Mon Oct 27, 2008 10:32 am
Thanks so much to both of you for the two approaches. :D
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