Find the sum of 24 terms of the series 2(1/2), 3(1/3), 4(1/6), 5 ......?

Question

Find the sum of 24 terms of the series \(2\frac{1}{2}\), \(3\frac{1}{3}\), \(4\frac{1}{6}\), 5, ......? 

(a) 200 

(b) 185 

(c) 290 

(d) 250

Answer 1

Answer : (c) 290

Here a = \(\frac{5}{2}\), d =  \(\frac{10}{3}\) - \(\frac{5}{2}\)  = \(\frac{20-15}{6}\) = \(\frac{5}{6}\)

∴ S₂₄  = \(\frac{24}{2}[2 \times \frac{5}{2} +(24 - 1) \frac{5}{6}]\)   \(\big[\because S_n = \frac{n}{2} [2a + (n-1)d]\big]\)

= 12 [5 + \(\frac{23 \times 5}{6}\)] = 2 [30 + 115]

= 2 × 145 = 290.