The sum of the 5 th and the 7 th terms

Let the first term and common difference of AP are a and d, respectively.

According to the question,

$a_{5}+a_{7}=52$ and $a_{10}=46$

$\Rightarrow \quad a+(5-1) d+a+(7-1) d=52 \quad\left[\because a_{n}=a+(n-1) d\right]$

and $\quad a+(10-1) d=46$

$\Rightarrow \quad a+4 d+a+6 d=52$

and $\quad a+9 a=46$

$\Rightarrow \quad 2 a+10 d=52$

and $a+9 d=46$

$\Rightarrow$ $a+5 d=26$ ...(i)

$a+9 d=46$ ....(ii)

On subtracting Eq. (i) from Eq. (ii), we get

$4 d=20 \Rightarrow d=5$

From Eq. (i), $a=26-5(5)=1$

So, required AP is $a, a+d, a+2 d, a+3 d, \ldots$ i.e., $1,1+5,1+2(5), 1+3(5), \ldots$ i.e., $1,6,11,16, \ldots$