For the A.P. : 3, \(5\frac{1}{2}\), 8, .... a = 3, d = \(2\frac{1}{2}\), number of terms = n
⇒ Sn = \(\frac{n}{2}\) [2a + (n – 1) d] = \(\frac{n}{2}\) ( 6 + (n -1) \(\times\) \(\frac{5}{2}\)) ... (i)
For the A.P.: \(16\frac{1}{2}\), \(28\frac{1}{2}\), \(40\frac{1}{2}\) ... a = \(16\frac{1}{2}\), d = 12.
∴ T2n= a + (2n – 1) d = \(16\frac{1}{2}\) + (2n - 1) \(\times\)12 ... (ii)
Given, Sn = T2n ⇒ \(\frac{n}{2}\)(6 +(n-1)\(\frac{5}{2}\)) = \(\frac{33}{2}\)+ (2n -1) \(\times\)12 ... [From (i) and (ii)]
⇒ Neglecting – ve value, we have n = 18.