The sum of the first n terms of the arithmetical progression 3, 5(1/2) , 8, ... is equal to the 2nth term of the

Question

The sum of the first n terms of the arithmetical progression 3, \(5\frac{1}{2}\) , 8, ... is equal to the 2nth term of the A.P. \(16\frac{1}{2}\), \(28\frac{1}{2}\), \(40\frac{1}{2}\),  .... Calculate the value of n.

Answer 1

For the A.P. : 3, \(5\frac{1}{2}\), 8, ....  a = 3, d = \(2\frac{1}{2}\), number of terms = n

⇒ S_n = \(\frac{n}{2}\) [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. 

∴ T_2n= a + (2n – 1) d = \(16\frac{1}{2}\) + (2n - 1) \(\times\)12 ... (ii) 

Given, S_n = T_2n ⇒ \(\frac{n}{2}\)(6 +(n-1)\(\frac{5}{2}\)) = \(\frac{33}{2}\)+ (2n -1) \(\times\)12 ... [From (i) and (ii)]

⇒ Neglecting – ve value, we have n = 18.