Question

How many terms of the A.P. : 24, 21, 18, ................ must be taken so that their sum is 78?

Answer 1

Let the number of terms taken be n.

The given A.P. is 24, 21, 18, ................

Here, a = 24 and d = 21 – 24 = –3

`S = n/2 [2a + (n - 1)d]`

`=> 78 = n/2 [2 xx 24 + (n - 1)xx (-3)]`

`=> 78 = n/2 [48 - 3n + 3]`

`=>` 156 = n[51 – 3n]

`=>` 156 = 51n – 3n²

`=>` 3n² – 51n + 156 = 0

`=>` n² – 17n + 52 = 0

`=>` n² – 13n – 4n + 52 = 0

`=>` n(n – 13) – 4(n – 13) = 0

`=>` (n – 13)(n – 4) = 0

`=>` n = 13 or n = 4

∴ Required number of terms = 4 or 13

Answer 2

Given: 24, 21, 18, ........ are in A.P.

Here, a = 24, d = 21 – 24 = – 3

Sum, `S_n = n/2 [2a + (n - 1)d]`

⇒ `78 = n/2 [2 xx 24 + (n - 1)(-3)]`

⇒ 156 = n(48 – 3n + 3)

⇒ 156 = n(51 – 3n)

⇒ 3n² – 51n + 156 = 0

⇒ 3n² – 12n – 39n + 156 = 0

⇒ 3n(n – 4) – 39(n – 4) = 0

⇒ (n – 4)(3n – 39) = 0

∴ (n – 4) = 0 and (3n – 39) = 0

⇒ n = 4 and n = 13

When n = 4,

`S_4 = 4/2 [2 xx 24 + (4 - 1)(-3)]`

= 2(48 – 9)

= 2 × 39

= 78

When n = 13

`S_13 = 13/2 [2 xx 24 + (13 - 1)(-3)]`

= `13/2 [48 + (-36)]`

= `13/2 xx 12`

= 78

Hence, the number of terms is n = 4 or n = 13.