4oops ... I didn't consider them
. . . .
Q1 Suppose 0.27 ,x ,0.72 are in GP.If x is a positive irrational no. and S is the sum of first four terms of such GP.Find S
Q2 Let a_n=111....1.Find no. of primes in sequence a2,a3,a4,a5,a6
n times
Q3 Let Let a_n=111....1.Find no. of perfect squares in sequence
n times
a2,a3.....a100
Q4 Let a_n=111....1.Find remainder when a124 divided by 271
n times
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24 Answers
62Bipin, the question asks only uptil a6 :P
i have my doubts if all these will be composites (Gut feeling)
there could be one way to prove that
10n-1 has a factor other than 3 when n is not a multiple of 3
but i cant think of that either
could not apply Fermat's convincingly
1For Q2 : an is always composite whenever n is composite
however I couldn't figure out when n is prime
but here upto a6 it is easy to check
62in 1, you have found the common ratio and the first term..
what s there to now understand :O
2 is solved in #18
111111 = 41 x 271
a124 = 11111 x 10119 + 11111 x 10114 + ...... + 11111 x 104 + 1111
all the terms in expansion of a124 are divisible by 271 except 1111
so remainder will be same as when 1111 is divided by 271 which is 27
answer is 27
62I din see that we dont have to prove it [3]
infact i dont find a reason why they should have used that phrase
24also explain "so we have to prove that √24/121 is irrational which is true" in post #15
62Q2 Let .Find no. of primes in sequence a2,a3,a4,a5,a6
n times
This one is a sitter.. i din notice that this ended at a6 :P
I thought we have to find this for a general n
for n=2, 11, 111, 1111, 11111, 111111
we can say easily that a3, a4, a6 are not prime
a2 =11 is a prime
we just need to check for 11111 whcih is not a prime ( I think there should be a nice logic using some congruences to prove this but am not able to give one right away)
62tushar for 3 , i did not understand your final arguement
the actual proof that i know is this
11, 111, 1111, 11111, and so on are always of the form of 4n+3
so they are never perfect squares.
the only perfect square in this sequence is 1 itself
62. . . .
Q1 Suppose 0.27 ,x ,0.72 are in GP.If x is a positive irrational no. and S is the sum of first four terms of such GP.Find S
dots show repetition like
.
0.1 = .111111111111111111111111111111111111 (infinitely many times)
. .
0.27 = 27/99 = 3/11
and
. .
0.72 = 72/99 = 8/11
so we have to prove that √24/121 is irrational which is true
41) If I understood the qs properly
then 0.27,x,0.72 r in G P
So x = 0.18 √6
S = 0.27 + 0.18√6 + 0.72 + 0.72x 2/3 x √6
11@ Soumya
Ne need to say sorry, my solution is " incomplete" .
Your solution is perfect [1]
1@ tushar -- extremely sorry but that is the whole problem , to prove 10n+1 - 1 is not a perfect square at all .
11111 ..... n times 1 = 100 + 101 ............... +10n =
=1 ( 10n+1 -1 ) /10-1
= 10n+1 -1 / 9
We shall show that 10n+1 - 1 cannot be a perfect square .
If possible , let k2 = 10n+1 -1
or k2 + 1 - 10n+1 =0
for positive solutions , discriminant >= 0
Hence 1 - 4 . 10n+1 > =0
which is not possible for any permitted n.
Hence the result .
11Don't know if I am correct or not,
Ans 3) Tn = 11111111.......1 (n+1 times)
= 1 + 10 + 102 + 10 3 +...........+10 n
= 1 (10 n+1 -1)(10-1) = 10 n+1 -1 9
which is not a perfect square for all n belonging to Natural number.