Question

How many terms of series 24 + 20 + 16 + ……. must be taken so that their sum is 72. 

Answer 1

In the given A.P; a = 24, d = 4, S_n = 72 

we have S_n = n/2{2a + (n – 1)d} 

72 = n/2{2(24) + (n – 1)(-4)} 

= 144 = n{48 – 4n + 4} 

⇒ 144 = n{52 – 4n} 

⇒ 144 = 52n – 4n² 

⇒ 4n² – 52n + 144 = 0 

⇒ n² – 13n + 36 = 0 

⇒ (n – 9)(n – 4) = 0 

∴ n = 9, 4