The question was:
An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first $11$ terms is $33$, find the fourth term.
I considered the first term to be $a$ and common difference to be $d$. So I used the formula for sum of arithmetic progressions $S_n=\frac n2[2a+(n-1)d]$ and found
$$\frac {11}{2}(2a+10d)=33$$
$$\implies a+5d=3$$
Now I saw that $a=d= \frac12$ quite clearly as that's the only way the 6th term could've could've $3$, as the AP is clearly
$\frac 12$, $1$, $\frac 32$, $2$, $\frac 52$, $3$ (6th term is $3$)
But what if I had got the equation of a term $a+nd= a_n$ such that $n$ is a much larger number? How would I have solved it that way?
Best Answer
To generalize the problem, consider an arithmetic progression in which the first term is a positive fraction, every alternate term is an integer, and the sum of the first $11$ terms is $S$.
If we take the convention that the general term of the progression is $a_k = a + kd,$ where $a_0 = a$ is the first term in the progression, then the conditions of the problem imply that $a_1$ and $a_3$ are integers. Therefore $a_3 - a_1 = 2d$ is also an integer, and therefore $d$ is a multiple of $\frac12.$
Depending on what we mean by "positive fraction," for a large enough $S$ the problem no longer has a unique solution. If "positive fraction" merely means a positive number that is not an integer, then if $S = 88$ then two possible solutions for the progression $a_k = a+kd$ starting at $a_0 = a$ are
$$a = \frac12, \ d = \frac32;$$ $$a = \frac{11}2, \ d = \frac12.$$
In general (under this interpretation) we will always have $a_5 = \frac S{11},$ and the only requirement for a solution for a given $S$ is that $\frac S{11} - 5d > 0$, which (given that $d$ is a multiple of $\frac12$) implies that $\frac S{11} - 5d \geq \frac12$ or $S - 55d \geq \frac{11}2$; if we let $d = m + \frac12$ for integer $m \geq 0$ (that is, $d$ is a multiple of $\frac12$ but not an integer), then $S - 55\left(m + \frac12\right) \geq \frac{11}2$, which is equivalent to $S \geq 33 + 55m.$ Clearly there is no solution if $S < 33,$ and if $S < 88$ there can be only one solution (in which $d = \frac12$).
Note that since $a_5$ is an integer, since $a = a_0 = a_5 - 5d,$ and since $d$ is a multiple of $\frac12,$ it follows that $a$ also is a multiple of $\frac12.$
If "positive fraction" means a number strictly between $0$ and $1,$ then the only possible value of $a$ is $a = \frac12$ and there is at most one solution.
Note that under either of the interpretations given above, if there is a solution at all, there is a solution in which $d = \frac12.$ Also, since $S = 11a_5,$ and $a_5$ is an integer, $S$ is a multiple of $11.$ And if $S \geq 33 + 55m$ for a non-negative integer $m,$ then there also are solutions in which $d$ can be any odd multiple of $\frac12$ from $\frac12$ itself up to $m + \frac12.$ That is, there are two solutions if $S = 88$ (shown above), three solutions if $S = 143$, and so forth.