Question -
Answer -
Let a, a + d, a + 2d,a + 3d, be the A.P.
Here a is the first term and d is the common difference
an = a + (n – 1) d
Now a7 = a + (7 – 1) d = a + 6d = 32 ….(i)
and a13 = a + (13 – 1) d = a + 12d = 62 ….(ii)
Subtracting (i) from (ii)
6d = 30
=> d = 5
a + 6 x 5 = 32
=> a + 30 = 32
=> a = 32 – 30 = 2
A.P. will be 2, 7, 12, 17, ………..