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The 1025th term in the sequence are 1,22...

The 1025th term in the sequence are `1,22,4444,88888888,"…"`is

A

`2^(9)`

B

NONE

C

`2^(11)`

D

`2^(12)`

Text Solution

Verified by Experts

The number of digits in each term of the sequence are `1,2,4,8,"…"` which are in GP. Let 1025th term is `2^(n)`.
Then `1+2+4+8+"..."+2^(n-1)lt1025 le1+2+4+8+"..."+2^(n)`
` implies ((2-1)(1+2+2^(2)+2^(3)+"..."+2^(n-1)))/((2-1))lt1025`
` le ((2-1)(1+2+2^(2)+2^(3)+"..."+2^(n)))/((2-1))`
` implies 2^(n)-1lt1025 le2^(n+1)-1 implies 2^(n)lt1026 le 2^(n+1)"....(i)"`
or `2^(n+1)ge 1026gt 1024`
`implies 2^(n+1)gt2^(10) implies n+1gt10`
` therefore ngt9 therefore =10` [which is always satisfy Eq. (i)]
Hence, (b) is the correct answer.
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