Question

8. Find the sum of first n terms of the series
(i) 7+77 + 777 + ..... (ii) 0.4 + 0.94 + 0.994 + ......

Answer 1

(i) 7 + 77 + 777 +......
= 7[ 1 + 11 + 111 + ....... ]
= 7/9 [ 9 + 99 + 999 + ....... ]
= 7/9[ (10 - 1) + (100 - 1) + (1000 - 1) + ...... ]
= 7/9[ (10 + 100 + 1000 + .......) - (1 + 1 + 1 + ......)]
Let there are n terms ,
Here, 10, 100, 1000 , ...... n terms are in GP
so, Sn = 10[10ⁿ - 1]/(10 - 1) = 10[10ⁿ - 1]/9
Now,
Sum of given series = 7/9[ 10(10ⁿ - 1)/9 - n ]

(ii)0.4 + 0.94 + 0.994 + ....... n terms
= [ (1 - 0.6) + (1 - 0.06) + (1 - 0.006) + ......... n terms ]
= [ (1 + 1 + 1 + 1 + .... n terms ) - (0.6 + 0.06 + 0.006 + 0.0006 + ..... n terms )]
Here, 0.6 + 0.06 + 0.006 + 0.0006 +....n terms Is the sum of GP .
Now, Sn = 0.6[1 - (0.1)ⁿ]/(1 - 0.1) = (2/3) [1 - (0.1)ⁿ]

Now, sum of given series = [ n - (2/3)(1 - 0.1ⁿ)]

Answer 2

( 1 ). 7 + 77 + 777 +......n

= 7[ 1 + 11 + 111 + .......n ]

= 7/9 [ 9 + 99 + 999 + ....... n]

= 7/9[ (10 - 1) + (100 - 1) + (1000 - 1) + ...... n]

= 7/9[ (10 + 100 + 1000 + .......n) - (1 + 1 + 1 + ......n)]

= 7/9[ 10[10ⁿ - 1]/(10 - 1) - n]

= 7/9[ 10(10ⁿ - 1)/9 - n ]


( 2 ). 0.4 + 0.94 + 0.994 + ....... n terms

= [ (1 - 0.6) + (1 - 0.06) + (1 - 0.006) + ......... n terms ]

= [ (1 + 1 + 1 + 1 + .... n terms ) - (0.6 + 0.06 + 0.006 + 0.0006 + ..... n terms )]


=[ n - 0.6[1 - (0.1)ⁿ]/(1 - 0.1) ]

= [ n - (2/3)(1 - 0.1ⁿ)]



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