Mu Sigma A.P, G.P, H.P Quiz-1

a=3, d= -3, n=10
we know ,n\'th term = a+(n-1)d
=3+(10-1)*(-3)
=3-27 = -24

First digit is 1;
2nd lind: 1 no. of 1 (in the first line)
3rd line: 1 no. of 2 and 2 nos. of 1 (in the 2nd line)
4th line: 1 no. of 1, 1 no. of 2 and 2 nos. of 1 (in the 3rd line)
5th line: 3 nos. of 1, 2 nos. of 2 and 2 nos. of 1 (in the 4th line)
6th line: 1 no. of 3, 1 no. of 1, 2 nos. of 2 and 2 nos. of 1 (in the fifth line)
.
.
.
.
.
similarly, the next line will be [31][13][12][11][13][12][21] = 31131211131221

Solution:
Option(C) is correct

Let the first installment be 'a' and the common difference between any two consecutive installments be 'd'

Using the formula for the sum of an A.P.

S=n/2[2a+(n−1)d]

We have,

3600=402[2a+(40−1)d]

⇒180=2a+39d-------- (i)

2400=302[2a+(30−1)d]

⇒160=2a+29d-------- (ii)

On solving both the equations we get:

d=2 and a=51

Let n be the number of members in the club

Then,

250=n/2[2×7+(n−1)×3/12]

n=25

Let the progression be a,a+d,a+2d .........

⇒a+11d=7a+7d
⇒6a=4d

⇒3a=2d
Also, a+7d=10a+3
⇒7d=9a+3
⇒7d=6d+3

⇒d=3,a=2
The GP is 2,2×3,2×32,2×33,2×34
So the 5th term is 162

The difference between 3-digit number and its reverse is 99 times the difference between its extreme (hundred and units) digits.

As the first difference is 396, the second is 4.

Further, as the digits are in AP and the hundred\'s digits is less than the unit\'s digit, we have following possibilities:

135,246,357,468,579.….

400+800+1200+......144000 (value after 24 years).

We know that, last term

=n/2[2a+(n−1)d]

Where, a is the first term and d= common difference between the term.

Then,

144000=24/2[2a+(24−1)×400]

⇒a= Rs 1400

Hence, value of the certificates after 13th year:

=1400+(13−1)×400
= Rs 6200

The number of members in A on Oct 2, 2007:

=n+4m=n+80n=81n
Number of members in B on Oct 2, 2007:

=n×p×p×p×p
As given in the question, now both organisations have the same number of members.

⇒81n=n×p^4
⇒p=3

24=42−k9‾√⇒k9‾√=18⇒k=6

0=42−6x√ (because maximum number of compartments that can be carried can be obtained by finding when the speed is zero and subtracting one from it).

42=6x√⇒x=49

∴ maximum number of compartments carried by the engine,

=49−1=48