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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.

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    $\begingroup$ This can't be answered for sure without more information. Is there anything else specified about this sequence? $\endgroup$
    – anon
    Jun 17, 2020 at 15:50
  • $\begingroup$ Hi @anon! Other than what is stated in my question, 'a' and 'd' are considered constants. The type of sequence is not specified. $\endgroup$
    – romeoPH
    Jun 17, 2020 at 15:54
  • $\begingroup$ 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. $\endgroup$
    – Andrei
    Jun 17, 2020 at 16:19
  • $\begingroup$ I'll suggest $42$. And what if $a$ equals $-4d$? $\endgroup$ Jun 17, 2020 at 17:04

1 Answer 1

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You cannot (uniquely) answer the question from the given information.

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.

However, there are infinitelymany sequences starting with the first four terms as given.
You could simply take your favorite sequence and replace the first four terms, for example take a constant sequence:
$x_0= \frac 1 a$, $x_1= \frac 1 {a+d}$, ..., $x_4= \frac 1 {a +3d}$, $x_5= 0$, $x_6=0$, $x_n=0$ for $n>4$

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