1st AP = 63, 65, 67, ...
Here, a = 63, d = 65 – 63 = 2
and 2nd AP = 3, 10, 17, ...
Here, a = 3, d = 10 – 3 = 7
According to the question,
63 + (n – 1)2 = 3 + (n – 1)7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n – 4
⇒ 65 = 7n – 2n
⇒ 5n = 65
⇒ n = 13
13th term of the given AP’s are same.
Now, we will find the 13th term
We have,
an = a + (n – 1)d
a13 = 63 + (13 – 1)2
a13 = 63 + 12 × 2
a13 = 63 + 24
a13 = 87