Let the first and common difference of AP are a and d, respectively.
According to the question,
a5+a7=52 and a10=46
⇒ a+(5−1)d+a+(7−1)d=52 [∵an=a+(n−1)d]
And a + (10 - 1)d = 46
⇒ a + 4d + a + 6d = 52
And a+ 9d = 46
⇒ 2a + 10d = 52
And a + 9d = 46
⇒ a + 5d = 26 . . . . . .(i)
a + 9d = 46 . . . . . . .(ii)
On subtracting eq.(i) from eq. (ii), we get;
4d = 20 ⇒ d = 5
From eq. (i), a = 26 - 5(5) = 1
So, required AP is a, a + d, a + 2d, a + 3d, . . . . i.e, 1, 1 + 5, 1 + 2(5) , 1 + 3(5), . . . . i.e, 1, 6, 11, 16,....