How many terms at most can be taken in sequence 1, 74, 894, 9688, 88997, 897999

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21 Aug 2009 Max possible terms in series

How many terms at most can be taken in sequence
1, 74, 894, 9688, 88997, 897999

OPtion

1. 10
2. 20
3. 8
4. 597
5. 1177
6. 25
7. 9
8. 10000000
9. 200000000
10. infinite

Solution

This sequence contains 1 digit extra in every next number.
1 | 74 | 894 | 9688 | 88997 | 897999
=> 1 | 7 + 4 |8 + 9 + 4| 9 + 6 + 8 + 8| 8 + 8 + 9 + 9 +7|8 + 9 + 7 + 9 + 9 + 9 |
=> 1| 11| 21 | 31 | 41 | 51| ….
In ninth term sum of all digit of the number will be 81.
After that it should be 91 in tenth term which is not possible because the greatest
number of ten digit is 9999999999, the sum of its all digits = 9 × 10 = 90

Correct Option: 7 Best Solution (1)

Deepak   14 years AGO

1 1 1
74 11 2
894 21 3
9688 31 4
88997 41 5
897999 51 6
8989999 61 7
89999999 71 8
999999999 81 9
See the sequence no more posibility of increasing a digit to make the sum of total digits 91,
so maximum possible terms in the sequence are 9.

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