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in Geometric Progressions by (15.7k points)

Find the sum of the series : 

NOTE: The following terms are not G.P. series, but we can convert them to form one. 

(i) 8 + 88 + 888 + …. To n terms 

(ii) 3 + 33 + 333 + …. To n terms 

(iii) 0.7 + 0.77 + 0.777 + …. To n terms

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1 Answer

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The expression can be rewritten as 

[Taking 8 as a common factor] 

8(1+ 11 + 111+ … to n terms) 

[Multiplying and dividing the expression by 9] 

= \(\frac{8}{9}\)(9 + 99+ 999 + … to n terms) 

=\(\frac{8}{9}\)((10-1) + (100-1) + (1000-1) + … to n terms ) 

=\(\frac{8}{9}\) ( ( 10 + 100 + 1000 + … to n terms) – ( 1+1+1+ … to n terms) 

= \(\frac{8}{9}\)( ( 10 + 100 + 1000 + … to n terms) – n) 

Sum of a G.P. series is represented by the formula, S\(a \frac{r^n - 1}{r-1}\), when r>1. 

‘Sn’ represents the sum of the G.P. series up to nth terms, 

‘a’ represents the first term, 

‘r’ represents the common ratio and 

‘n’ represents the number of terms. 

Here, 

a = 10 

r = (ratio between the n term and n-1 term) 10 

n terms

∴ The sum of the given expression is 

= \(\frac{8}{9}\)( ( 10 + 100 + 1000 + … to n terms) – n)

\(\frac{8}{9} \big( \frac{10^{n+1}-10}{9} - n \big)\)

(ii) The given expression can be rewritten as [taking 3 common ] 

= 3( 1+11+111+ …to n terms) [Multiplying and dividing the expression by 9 ] 

= \(\frac{3}{9}\)( 9+99+999+ … to n terms ) 

= \(\frac{3}{9}\)( (10-1) + (100-1) + (1000-1) + … to n terms ) 

= \(\frac{3}{9}\)( ( 10+100+1000+ …to n terms ) – (1+1+1+ … to n terms) ) 

= \(\frac{3}{9}\)( (10+100+1000+ to n terms) – n )

Sum of a G.P. series is represented by the formula, Sn = \(\frac{r^n - 1}{r-1}\) , when r>1. 

‘Sn’ represents the sum of the G.P. series up to nth terms, 

‘a’ represents the first term, 

‘r’ represents the common ratio and 

‘n’ represents the number of terms. 

Here,

a = 10 

r = (ratio between the n term and n-1 term) 10 

n terms

∴ The sum of the given expression is

= \(\frac{3}{9}\)( (10+100+1000+ to n terms) – n )

\(\frac{3}{9}\) \(\big(\frac{10^{n+1} - 10}{9} - n\big)\)

(iii) We can rewrite the expression as 

[taking 7 as a common factor] 

= 7(0.1+0.11+0.111+ … to n terms) 

[multiplying and dividing by 9 ] 

= \(\frac{7}{9}\)( 0.9+0.99+0.999+ … to n terms ) 

= \(\frac{7}{9}\)( (1-0.1)+(1-0.01)+(1-0.001)+ … to n terms) 

= \(\frac{7}{9}\)( (1+1+1+ … to n terms )–(0.1+0.01+0.001+… to n terms )) 

= \(\frac{7}{9}\)( n – (0.1+0.01+0.001+ … to n terms ) )

Sum of a G.P. series is represented by the formula, Sn = a\(\frac{1-r^n}{1-r}\), when |r|<1. 

‘Sn’ represents the sum of the G.P. series up to nth terms, 

‘a’ represents the first term, 

‘r’ represents the common ratio and 

‘n’ represents the number of terms. 

Here, 

a = 0.1 

r = (ratio between the n term and n-1 term) 0.1 

n terms

[multiplying both numerator and denominator by 10]

⇒ Sn = \(\frac{1-0.1^n}{9}\)

∴ The sum of the given expression is 

= \(\frac{7}{9}\)( n – (0.1+0.01+0.001+ … to n terms ) ) 

= \(\frac{7}{9}\)( n – (\(\frac{1-0.1^n}{9}\)) )

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