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, Sn = 72

we have Sn = 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