@vbhvgupta said:Q3
1/64?
a=3(ar+ar^2+......)
4=3*4(r+r^2+......)
1/3=(r/1-r)
r=1/4
fifth term = a*r^4=(4)(1/4)^4=1/64
The set of the three digits of such a number can be arranged
to form an increasing arithmetic sequence. There are 8 possible sequences
with a common diference of 1, since the Żrst term can be any of the digits
0 through 7. There are 6 possible sequences with a common diference of
2, 4 with a common diference of 3, and 2 with a common diference of
to form an increasing arithmetic sequence. There are 8 possible sequences
with a common diference of 1, since the Żrst term can be any of the digits
0 through 7. There are 6 possible sequences with a common diference of
2, 4 with a common diference of 3, and 2 with a common diference of
4. Hence there are 20 possible arithmetic sequences. Each of the 4 sets
that contain 0 can be arranged to form 2 *2! = 4 diferent numbers, and
the 16 sets that do not contain 0 can be arranged to form 3! = 6 diferent
numbers. Thus there are a total of 4 * 4 + 16 * 6 = 112 numbers with the
required properties.
that contain 0 can be arranged to form 2 *2! = 4 diferent numbers, and
the 16 sets that do not contain 0 can be arranged to form 3! = 6 diferent
numbers. Thus there are a total of 4 * 4 + 16 * 6 = 112 numbers with the
required properties.
@raopradeep
there is the doubt sign change happens between 0&2 and
4&5
F(4) = +45
F(5) = +35
its all clear ...
there is the doubt sign change happens between 0&2 and
4&5
F(4) = +45
F(5) = +35
its all clear ...
Q5
@mailtoankit said:option c?
@saurav5517 said:C N+2 divisible by 6..??
@ishu1991 said:N+2 is divisible by 6
explain.
@vbhvgupta said:explain.
took N=4
sum = 1+2+3+4 = 10
sum of squares = 1^2+2^2+3^2+4^2 = 30
so N+2 = 4+2 = 6 divisible by 6
@vbhvgupta said:explain.
A(N)= N*(N+1)/2
B(N)= N*(N+1)*(2N+1)/6
Now its given B(N) is a mutiple of A(N) ..let the multiple be k which shall be an integer
So.. B(N) = k*A(N)
N*(N+1)*(2N+1)/6 = K * N(N+1)/2
2N+1 = 3K.
When you take cases you shall conclude that N+2 must be divisible by 6 ..
Ram and Shyam are moving along two straight roads that intersect perpendicularly at X.When ram is at X,Shyam is 500m from X.After 2 mins both Ram and Shyam are at equal distance fro m X,and after 8 more mins they are again at equal distance from X.What is the ratio of Ram's speed to Shyam's speed?
a>2:3 b>4:5 c>5:8 d>1:2 e>None
Guys plz help..
Q6