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Question

What is the sum of the series 0.5+0.55+0.555+.... to n terms?

A
59[n29(1110n)]
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B
19[529(1110n)]
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C
19[n59(1110n)]
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D
59[n19(1110n)]
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Solution

The correct option is D 59[n19(1110n)]
0.5+0.55+0.555+............nterms
=5{0.1+0.11+0.111+........}
=59(910+99100+9991000+..........)
=59(1110+11100+111000+....... n terms)
=59(n(110+1100+11000+......... n terms))
Now, 110,1100,11000 ,.... are in GP
0.5+0.55+0.555+............n terms
=59(n(110(1(110)n)1110))
=59[n19(1110n)]

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