How many terms of the sequence √3, 3, 3√3,… must be taken to m

1.	How many terms of the sequence √3, 3, 3√3,… must be taken to make the sum 39 + 13√3 ?
Answer» Given: Sum of GP = 39 + 13√3 Where, a =√3, r = 3/√3 = √3, n = ? By using the formula, Sum of GP for n terms = a(rⁿ – 1 )/(r – 1) 39 + 13√3 = √3 (√3ⁿ – 1)/ (√3 – 1) (39 + 13√3) (√3 – 1) = √3 (√3ⁿ – 1) Let us simplify we get, 39√3 – 39 + 13(3) – 13√3 = √3 (√3ⁿ – 1) 39√3 – 39 + 39 – 13√3 = √3 (√3ⁿ – 1) 39√3 – 39 + 39 – 13√3 = √3ⁿ⁺¹ – √3 26√3 + √3 = √3ⁿ⁺¹ 27√3 = √3ⁿ⁺¹ √3⁶ √3 = √3ⁿ⁺¹ 6+1 = n + 1 7 = n + 1 7 – 1 = n 6 = n ∴ 6 terms are required to make a sum of 39 + 13√3

How many terms of the sequence √3, 3, 3√3,… must be taken to make the sum 39 + 13√3 ?

Answer»

Given:

Sum of GP = 39 + 13√3

Where, a =√3, r = 3/√3 = √3, n = ?

By using the formula,

Sum of GP for n terms = a(rⁿ – 1 )/(r – 1)

39 + 13√3 = √3 (√3ⁿ – 1)/ (√3 – 1)

(39 + 13√3) (√3 – 1) = √3 (√3ⁿ – 1)

Let us simplify we get,

39√3 – 39 + 13(3) – 13√3 = √3 (√3ⁿ – 1)

39√3 – 39 + 39 – 13√3 = √3 (√3ⁿ – 1)

39√3 – 39 + 39 – 13√3 = √3ⁿ⁺¹ – √3

26√3 + √3 = √3ⁿ⁺¹

27√3 = √3ⁿ⁺¹

√3⁶ √3 = √3ⁿ⁺¹

6+1 = n + 1

7 = n + 1

7 – 1 = n

6 = n

∴ 6 terms are required to make a sum of 39 + 13√3