Question

Find the sum of the series, 1+√3+3+3√3+....81√3

Answer 1

Answer:

S(10) = 121(√3 + 1)

Note:

★ A sequence is which the common ratio between the consecutive terms are equal is called GP ( Geometric progression ) .

★ Sum of the terms of a sequence is called series .

★ Sum of the terms of a GP is called Geometric series .

★ The nth term of a GP is given by ;

a(n) OR T(n) = ar^(n - 1) , where a is the first term and r is the common ratio of the GP .

★ The general form of a GP is ;

a , ar , ar² , ar³ , . . . , ar^(n - 1)

★ In a GP , the common ratio is given by;

r = a(n) / a(n-1) .

a2/a1 = a3/a2 = a4/a3 = ... = a(n)/a(n-1) = r

★ If a , b , c are in GP , then b² = ac

★ The sum of n terms of a GP is given as ;

S(n) = a×(rⁿ - 1) / (r - 1)

Solution:

The given series is ;

1 + √3 + 3 + 3√3 + . . . + 81√3

Here,

First term , a = 1

Common ratio , r = a2/a1 = √3/1 = √3

nth term , a(n) = 81√3

Now,

=> a(n) = 81√3

=> a×r^(n - 1) = 81√3

=> 1×(√3)^(n - 1) = 81√3

=> √3^(n - 1) = √3^9

=> n - 1 = 9

=> n = 9 + 1

=> n = 10

Now,

The sum of the 10 terms of the given geometric series will be given as ;

=> S(n) = a×(rⁿ - 1) / (r - 1)

=> S(10) = 1×(√3¹°–1) / (√3 - 1)

=> S(10) = (243 - 1) / (√3 - 1)

=> S(10) = 242 / (√3 - 1)

=> S(10) = 242(√3 + 1) / (√3 - 1)(√3 + 1)

=> S(10) = 242(√3 + 1) / (√3² - 1²)

=> S(10) = 242(√3 + 1) / (3 - 1)

=> S(10) = 242(√3 + 1) / 2

=> S(10) = 121(√3 + 1)

Hence ,

S(10) = 121(√3 + 1)