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

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Best 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

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