Arithmetic Sequence

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by Geva@EconomistGMAT » Sun Nov 28, 2010 3:11 am
What do we know:

t1=23.
tn=t(n-1)-3 . In words: each term is equal to the previous term minus 3, or "each term is 3 less than the previous term".
From here, just list the terms until you reach -4:

t1=23
t2 = 23-3=20
t3 = 20-3=17
t4=14
t5=11
t6=8
t7=5
t8=2
t9=-1
t10=-4

Thus, the n where the term tn equals -4 is 10.
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by beat_gmat_09 » Sun Nov 28, 2010 3:14 am
t1 = 23 , each is decremented by 3
After 9 terms (i.e. n will be 1+9 = 10) t10 will be = 23 - 9*3 = 23-27 = -4
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by rishab1988 » Sun Nov 28, 2010 3:53 am
My approach.

You can see that the question mentions that it is A.P.You can also see that Common difference =-3

To find which term =-4

Let's assume -4 is the last term
First term =23 [first term]

The formula for no of terms = (First term- Last term)/distance between each term = (23+4)/3=9 [ distance is always +ve]

Now add 1 for -4 and 23 are both included in the sequence.No of terms =10

[ Think of this as "How many terms are there from 1 to 2? -> (First - second term)/distance bt each term = (2-1)/1 = 1.Since we know that both 1 and 2 are in the sequence,add 1 -> no of terms =2]

So if t(1)=23 then t(10) =-4

Hence answer =10

[This procedure saved me from tedious calculation]

if you wanna know the high school A.P forumula for no terms :

An = Ao + (n-1)d [ An is the nth term Ao is the first term d is the common difference n is the no of terms]

We know that:

Ao=23
An=-4
d=-3

-4 = 23 + (n-1)(-3)
-27= -3n+3
n=10

Therefore the n in An =-4 is 10

If you wanna know the high school formula for sum :

Sum = n/2 [2a+(n-1)d] [ n is the no of terms.a is the first term d is the common difference]

Also in an A.P avg = (First term value + last term value)/2 = Avg

If you know the no of terms, you can calculate sum directly from = Avg* no terms