Advertisement Remove all ads

Advertisement Remove all ads

Advertisement Remove all ads

Sum

Find the value of n, if the sum to n terms of the series `sqrt(3) + sqrt(75) + sqrt(243) + ......` is `435 sqrt(3)`

Advertisement Remove all ads

#### Solution

t_{1} = `sqrt(3)`

t_{2} = `sqrt(75)`

= `sqrt(25 xx 3)`

= `5sqrt(3)`

t_{3} = `sqrt(243)`

= `sqrt(81 xx 3)`

= `9sqrt(3)`

Here t_{1} = `sqrt(3)`

t_{2} = `5sqrt(3)`

t_{3} = `9sqrt(3)`

(i.e) a = `sqrt(3)`

d = `5sqrt(3) - sqrt(3)`

= `4sqrt(3)`

S_{n} = `"n"/2[2"a" + ("n" - 1)"d"]`

= `435 sqrt(3)` ......(Given)

⇒ `"n"/2 [2sqrt(3) + ("n" - 1)4sqrt(3)] = 435sqrt(3)`

⇒ `("n" sqrt(3))/2 [2 + 4"n" - 4] = 435 sqrt(3)`

⇒ n[4n – 2] = 870

4n^{2} – 2n – 870 = 0

(÷ by 2)2n^{2} – n – 435 = 0

2n^{2} – 30n + 29n – 435 = 0

⇒ 2n(n – 15) + 29(n – 15) = 0

(2n + 29)(n – 15) = 0

⇒ n = `(-29)/2` or 15

n = `(-29)/2` not possible,

So n = 15

Concept: Finite Series

Is there an error in this question or solution?