For any sequence problems I usually look at the patterns it is creating by taking a couple of examples.

For (n = 1): t1 = 23

t2 = t(n-1) -3 = t(2-1) -3 = t(1) -3 = 23-3 = 20

t3 = t(n-1) - 3 = t(3-1) - 3 = t(2) - 3 = 20-3 = 17

So essentially for every successive t, the number decreases by 3

you want your final number to be -4

so 23 - x = -4

x = 27

27/3 = 9

n will have moved 9 down to give the value of -4

t2 to t10 is 9 down: so the value of n is 10.

Is this approach right?, we dont' need to use t1=23???


sangeethai,
u can take tn = -4 as the first term of the series. now use the formula -
tn = a + (n - 1) * d
where tn = -4, a = -4 and d = 3 , u can get n = 10.
_________________
Correct me If I am wrong


Regards,

Amitava

We know that this is an arithmetic progression with the common difference of -3.
i.e t1 = 23, t2 = 20, t3 = 17 .....
tn = t1 + (n-1)(-3) => 23 -3(n) + 3
or -4 = 26 - 3(n)
3(n) = 30 which gives n = 10.