Question -
Answer -
(i) 5 + 55 + 555 + … to nterms.
Let us take 5 as acommon term so we get,
5 [1 + 11 + 111 + … nterms]
Now multiply anddivide 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 nterms.
Let us take 7 as acommon term so we get,
7 [1 + 11 + 111 + … ton terms]
Now multiply and divideby 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 inG.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 nterms.
The given terms can bewritten 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 acommon term so we get,
5(0.1 + 0.11 + 0.111 +…n terms)
Now multiply anddivide 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 acommon term so we get,
6(0.1 + 0.11 + 0.111 +…n terms)
Now multiply anddivide 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)]