kashewman wrote:Hey all,

I consistently get sequence problems wrong, is there an easy fix? One problem I recently missed..

In the arithmetic sequence t(1) [all numbers are 'under' the t not an exponent], t(2),t(3)...t(n).., t(1) = 23 and t(n) = t(n-1) - 3 for each n>1. What is the value of n when t(n) = -4?

Any help is apprecaited, thanks!

You can easily solve this by going step by step, pluging in previous ( t(n-1) ) value

we know,
t(1) = 23
t(2) = t(1) - 3 = 20
t(3) = 17
4 = 14
5 = 11
6 = 8
7 = 5
8 = 2
9 = -1
10 = -4

However there are Two Arithment Progression equations you may use them, whenever you come across Arithmetic progression Problems..

in a sequence of number

a, a+d, a+2d..... l (where l is the last number)

Last number of the sequence is determined using following equation

Last number l = a + (n-1)d

And sum of all the numbers in the sequence can be determined using the following equation

SUM = (n/2)(2a + (n-1)d)

WHERE
n = number of numbers in the sequence
a = first number in the sequence
l = last numbr in the sequence
d = common difference

generally.. you will have 3 of these values and will be asked the fourth one and SUM Or sum will be given and asked to find n or something like that.

Hope this helps!!

I just figured these out and needed to have someone show me how to do it.

The key to success with these is to understand what is happening in the sequence and ten figure out where you stand and then just plug in the numbers in th esequence in order to come to the answer. The explanation in from the writer above did the exact same way I would approach the situation

kulksnikhil wrote:

kashewman wrote:Hey all,

I consistently get sequence problems wrong, is there an easy fix? One problem I recently missed..

In the arithmetic sequence t(1) [all numbers are 'under' the t not an exponent], t(2),t(3)...t(n).., t(1) = 23 and t(n) = t(n-1) - 3 for each n>1. What is the value of n when t(n) = -4?

Any help is apprecaited, thanks!

You can easily solve this by going step by step, pluging in previous ( t(n-1) ) value

we know,
t(1) = 23
t(2) = t(1) - 3 = 20
t(3) = 17
4 = 14
5 = 11
6 = 8
7 = 5
8 = 2
9 = -1
10 = -4

However there are Two Arithment Progression equations you may use them, whenever you come across Arithmetic progression Problems..

in a sequence of number

a, a+d, a+2d..... l (where l is the last number)

Last number of the sequence is determined using following equation

Last number l = a + (n-1)d

And sum of all the numbers in the sequence can be determined using the following equation

SUM = (n/2)(2a + (n-1)d)

WHERE
n = number of numbers in the sequence
a = first number in the sequence
l = last numbr in the sequence
d = common difference

generally.. you will have 3 of these values and will be asked the fourth one and SUM Or sum will be given and asked to find n or something like that.

Hope this helps!!

How Can we use these equations for the question above I am getting confused?