For positive integers $a,b$ and $k$ , define $$p(a,b,k):=\prod_{j=1}^k (a(j-1)+b)$$ That is the product of the first $k$ numbers in the arithmetic progression $an+b$ starting with $b$
Can $p(a,b,k)$ be a palindrome (in base $10$) for $k\ge 6$ ? Are there more than $2$ palindromes for $k= 5$ ?
In the case $k=4$ , there are still many palindromes.
For $a,b\le 10^4$ , they are summarized in this table :
4 [7, 9, 11, 13] 9009
4 [12, 14, 16, 18] 48384
4 [1, 5, 9, 13] 585
4 [91, 101, 111, 121] 123444321
4 [11, 26, 41, 56] 656656
4 [26, 42, 58, 74] 4686864
4 [1023, 1073, 1123, 1173] 1445949495441
4 [9901, 10001, 10101, 10201] 10203040404030201
4 [127, 245, 363, 481] 5432772345
4 [1059, 1407, 1755, 2103] 5499287829945
4 [471, 961, 1451, 1941] 1274785874721
4 [55, 583, 1111, 1639] 58388088385
4 [13, 543, 1073, 1603] 12141614121
4 [1864, 2428, 2992, 3556] 48152399325184
4 [8841, 9497, 10153, 10809] 9214414224144129
4 [77, 1111, 2145, 3179] 583341143385
But for $5\le k\le 20$ and the same range for $a,b$ , there are just the following two solutions :
5 [31, 33, 35, 37, 39] 51666615
5 [182, 362, 542, 722, 902] 23255355355232
For $k\ge 5$ , the following can be said about the difference $a$ :
If $a$ is coprime to $10$ , one of the factors of the product must be divisible by $5$ and the product is even, hence divisible by $10$. A palindrome is hence impossible.
If $a$ is even , but not divisible by $5$ , one of the factors must be divisible by $5$ , hence the only possible palindrome is $5\cdots 5$, in particular all factors must be odd.
But that's it. I found no more useful necessary conditions for a palindrome.