Question

Find the sum of the following series :
(i) 5 + 55 + 555 + … to n terms.

(ii) 7 + 77 + 777 + … to n terms.

(iii) 9 + 99 + 999 + … to n terms.

(iv) 0.5 + 0.55 + 0.555 + …. to n terms

(v) 0.6 + 0.66 + 0.666 + …. to n terms.

Answer 1

(i) 5 + 55 + 555 + … to n terms.

Let us take 5 as a common term so we get,

5 [1 + 11 + 111 + … n terms]

Now multiply and divide by 9 we get,

5/9 [9 + 99 + 999 + … n terms]

5/9 [(10 – 1) + (10² – 1) + (10³ – 1) + … n terms]

5/9 [(10 + 10² + 10³ + … n terms) – n]

So the G.P is

5/9 [(10 + 10² + 10³ + … n terms) – n]

By using the formula,

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

Where, a = 10, r = 10²/10 = 10, n = n

a(rⁿ – 1 )/(r – 1) =

(ii) 7 + 77 + 777 + … to n terms.

Let us take 7 as a common term so we get,

7 [1 + 11 + 111 + … to n terms]

Now multiply and divide by 9 we get,

7/9 [9 + 99 + 999 + … n terms]

7/9 [(10 – 1) + (10² – 1) + (10³ – 1) + … + (10ⁿ – 1)]

7/9 [(10 + 10² + 10³ + … +10ⁿ)] – 7/9 [(1 + 1 + 1 + … to n terms)]

So the terms are in G.P

Where, a = 10, r = 10²/10 = 10, n = n

By using the formula,

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

7/9 [10 (10ⁿ – 1)/(10-1)] – n

7/9 [10/9 (10ⁿ – 1) – n]

7/81 [10 (10ⁿ – 1) – n]

7/81 (10ⁿ⁺¹ – 9n – 10)

(iii) 9 + 99 + 999 + … to n terms.

The given terms can be written as

(10 – 1) + (100 – 1) + (1000 – 1) + … + n terms

(10 + 10² + 10³ + … n terms) – n

By using the formula,

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

Where, a = 10, r = 10, n = n

a(rⁿ – 1 )/(r – 1) = [10 (10ⁿ – 1)/(10-1)] – n

= 10/9 (10ⁿ – 1) – n

= 1/9 [10ⁿ⁺¹ – 10 – 9n]

= 1/9 [10ⁿ⁺¹ – 9n – 10]

(iv) 0.5 + 0.55 + 0.555 + …. to n terms

Let us take 5 as a common term so we get,

5(0.1 + 0.11 + 0.111 + …n terms)

Now multiply and divide by 9 we get,

5/9 [0.9 + 0.99 + 0.999 + …+ to n terms]

5/9 [9/10 + 9/100 + 9/1000 + … + n terms]

This can be written as

5/9 [(1 – 1/10) + (1 – 1/100) + (1 – 1/1000) + … + n terms]

5/9 [n – {1/10 + 1/10² + 1/10³ + … + n terms}]

5/9 [n – 1/10 {1 - (1/10)ⁿ}/{1 – 1/10}]

5/9 [n – 1/9 (1 – 1/10ⁿ)]

(v) 0.6 + 0.66 + 0.666 + …. to n terms.

Let us take 6 as a common term so we get,

6(0.1 + 0.11 + 0.111 + …n terms)

Now multiply and divide by 9 we get,

6/9 [0.9 + 0.99 + 0.999 + …+ n terms]

6/9 [9/10 + 9/100 + 9/1000 + …+ n terms]

This can be written as

6/9 [(1 – 1/10) + (1 – 1/100) + (1 – 1/1000) + … + n terms]

6/9 [n – {1/10 + 1/10² + 1/10³ + … + n terms}]

6/9 [n – 1/10 {1 - (1/10)ⁿ}/{1 – 1/10}]

6/9 [n – 1/9 (1 – 1/10ⁿ)]