# The 1st term is $\frac{1}{a}$, 2nd term is $\frac{1}{a+d}$, 3rd term is $\frac{1}{a+2d}$. Find the 5th term of the sequence?

If for some number a and d,if first term is 1⁄a, second term is 1/(a+d) ,thrid term is 1/(a+2d) and so on, then 5th term of the sequence is :________?

I am attempting to answer the question above. I assumed that I would just add a 'd' for every term, so I did the following:

First term: $$\frac{1}{a}$$

Second term: $$\frac{1}{a+d}$$

Third term: $$\frac{1}{a+2d}$$

Fourth term: $$\frac{1}{a+3d}$$

Fifth term: $$\frac{1}{a+4d}$$

My answer is $$\frac{1}{a+4d}$$; however, I worry that merely adding a 'd' might be wrong.

– anon
Jun 17, 2020 at 15:50
• Hi @anon! Other than what is stated in my question, 'a' and 'd' are considered constants. The type of sequence is not specified. Jun 17, 2020 at 15:54
• The "and so on" is ambiguous. Your sequence might be the form $1/(a+c_n d)$, where $c_0=0, c_1=1, c_2=2$. You assume that $c_3=3$, but that's not guaranteed. Let's choose $c_{n+1}=c_n^2+1$, with $c_0=0$. Then $c_1=0^2+1=1$, $c_2=1^2+1=2$, $c_3=2^2+1=5$, $c_4=5^2+1=26$. This is as valid sequence as the one you gave. Jun 17, 2020 at 16:19
• I'll suggest $42$. And what if $a$ equals $-4d$? Jun 17, 2020 at 17:04

If the sequence is the sequence $$(x_n)_n$$ defined by $$x_n=\frac 1 {a+(n-1)d}$$, then the fifth term is as you said.
$$x_0= \frac 1 a$$, $$x_1= \frac 1 {a+d}$$, ..., $$x_4= \frac 1 {a +3d}$$, $$x_5= 0$$, $$x_6=0$$, $$x_n=0$$ for $$n>4$$