number system
Q1. Let S(n) denote the sum of the digits if a natural number ‘n’.
If 2n + S(n) = 1998, then S(n) can be :
a. 24 b. 21 c. 17 d. Non of these
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Q2. Consider a sequence whose nth term {an } is given by
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{an} = 1/n!( n + 2) where n = 1, 2, 3. . . .
The sum of the first n terms of this sequence is given by
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Q3. An Arabian king celebrated his yth birthday in the year 2y2. In what year did he ascend to the throne if he was 17 years old at that time?
a. 1720 b. 1953 c. 1670 d. None of these
Q3
the year is 2y2 or 2y2 ?
if it is 2y2 then the ans is (c) 1670
and the value of ' y ' is 29 years
his date of birth is 1653.
thank u
shahid
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I Guess the question is
Q1. Let S(n) denote the sum of the digits of(not if) a natural number ‘n’.
If 2n + S(n) = 1998, then S(n) can be :
a. 24 b. 21 c. 17 d. Non of these
Ans:
The best way to solve such question is to refer to the available options
2n + S(n) = 1998
=>2n = 1998 - S(n)
Now S(n) cant be an odd number(in that case n 'll be fraction)
so the only number left is 24
thus 2n = 1998 - 24 = 1974
=> n = 987
and S(n) = 9 + 8 + 7 = 24 (ans a)
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YES u are correct.
I wasted lot of time on that question.
I request the persons who post the questions
to be more legible.
thank u
shahid
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Q2
{an} = 1/n!( n + 2) where n = 1, 2, 3. . . .
sum of the series is Sn= 1/3.1! + 1/4.2! + 1/5.3! + 1/6.4! + ------
=> 2/3! + 3/4! + 4/5! + 5/6! + ------
=>(3-1)/3! + (4-1)/4! + (5-1)/5! + (6-1)/6!+-----
=> 1/2 - 1/3! + 1/3! -1/4! +1/4! - -------
=> 1/2
thank u
shahid
n/a
n/a