The sum of the $5^{th}$ and the $7^{th}$ terms of an AP is 52 and the $10^{th}$ term is 46
then the AP is 1, 6, 11, 16.......
If true then enter $1$ and if false then enter $0$

Let $a$ be the first term and $d$ be the common difference of the given A.P.
Then,
$a_5+a_7=52$
$a_{10}=46$
$a+4d+a+6d=52$ and
$a+9d=46$
$\Rightarrow 2a+10d=\mathrm{52....}(1)$
$a+9d=\mathrm{46....}(2)$
multiplying (2) equation by 2 and we get,
$2a+18d=\mathrm{92....}(3)$
now subtract (1) from (3) then we get,
$8d=40$
$\Rightarrow d=5$
now put d in (1) equation we get
$a=1$
There for the given series is
$a,a+d,a+2d,a+3d...$
i.e. $1,6,11,\mathrm{16...}$