The ratio of sum of 2 AP's = (2a+ (n -1)d):(2A + (n-1)D) = (6+n):2
=> $$ a + \frac{n-1}{2}d\ :\ A+\frac{n-1}{2}D$$ = (6+n):2

=> observing L.H.S we can see that it is the ratio of the (n+1)/2 th terms of the A.P.s
Hence for the ratio of the 10th terms, we equate (n+1)/2 = 10 => n = 19
Hence ratio = (6+19)/2 = 25/2

Let the first term be a and the common difference be d. We have, (a+7d)/(a+16d) = 5/11.

Solving this, we get d=2a. Product of the first and sixth terms is $$a(a+5d) = a(a+10a) = 11a^2 = 176 => a=+- 4$$.
Since the series has only negative terms, a=-4 and d=-8.
Sum of the first 200 terms = (200/2)(2*-4 + 199*- 8) = -160000

Let the first term be a, second term be x and the third term be c.
Fourth term = a+c-x
Fifth term = a
Sixth term = x
In this way pattern will repeat.
26th term = 2nd term = x = 9
c = 5
2(a+c)*4+a+x = 58
9a+8c+x = 58
9a+40+9 = 58
a = 1

Initial height is 30m, after the first bounce, the height reached is 3/4(30). After the second bounce, the height reached = 3*3/4*4(30).

Total distance travelled = 30 + 3/4(30)*2 + 3/4(3/4(30))*2 + …

= 30 + 3/4(30)*2/(1-3/4) = 210m

If we consider the number carefully, then nth layer contains n(n+1)/2
$$\sum \frac{n(n+1)}{2} = 8436$$
$$\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}= 8436$$
n(n+1)(n+2) = 50616
n = 36

Let the first term of the AP be a and the common difference be d. So, the first term of the GP is a. Let the common ratio be r.
So, a + 2d = ar => d = a(r-1)/2
Also, $$a + 26d = ar^3$$
Solving the two equations, we get the values of r as 1, 3 and -4
Case 1: r = 1. In this case, all the terms of the GP will be the same. But, it is mentioned that the 70th term is greater than the 69th term. So, r cannot be 1
Case 2: r = -4. In this case, every odd term of the GP is positive and every even term is negative. So, the 69th term should be positive and the 70th term should be negative. So, $$T_{69} > T_{70}$$. But, this is not the case. So, the value of r cannot be -4.
r = 3 is the only possibility. So, required ratio = 25/729

given, 2b = a + c...(1)
also given, $$\frac{1}{b^2} - \frac{1}{a^2} = \frac{1}{c^2} - \frac{1}{b^2}$$
$$\frac{(a + b)(a - b)}{a^2b^2} = \frac{(b + c)(b - c)}{c^2b^2}$$
(b - a) = (c - b) as a,b,c are in A.P
so $$\frac{(a + b)}{a^2b^2} = \frac{(b + c)}{c^2b^2}$$
$$ac^2 + bc^2 = a^2b + a^2c$$
$$(ac^2 - a^2c) = (bc^2 - a^2b) = 0$$
ac(c - a) + b(c + a)(c - a) = (c-a)(ac + b(c + a)) = 0
=> c=a or ac + b(c + a) = 0 but c + a = 2b
thus a=b=c or $$ac + 2b^2$$ = 0

Since $$a\ne\ b\ne\ c$$ , $$ac + 2b^2$$ = 0

=> a,b,-c/2 are in GP.

This series is made up of 2 GPs. Consider the first term

2/3 = (3-1)/(1*3) = 1- 1/3

Consider the second term:

7/18 = (9-2)/(9*2) = 1/2 - 1/9

Similarly consider the third term,

23/108 = (27-4)/(27*4) = 1/4 - 1/27.

Consider the fourth term,

73/648 = (81-8)/(81*8) = 1/8 - 1/81

Now add all the terms together

1-(1/3)+(1/2)-(1/9)+(1/4)-(1/27)+…and so on

[1+(1/2)+(1/4)+(1/8)+….] -[(1/3)+(1/9)+(1/27)+(1/81)+…..]

= $$\frac{1}{1-\frac{1}{2}}-\frac{\frac{1}{3}}{1-\frac{1}{3}}$$

= 2-1/2 = 3/2

Let a, ar and ar$$^2$$ be the three sides of the right triangle.
Since, it is a right triangle, $$(ar^2)^2 = a^2 + (ar)^2$$
$$r^4 = r^2 + 1$$
$$r^4 - r^2 - 1 = 0$$
$$r^2 = (1 \pm \sqrt{1 - (-4)})/2$$
$$r^2 = (1 \pm \sqrt{5})/2$$
Since, $$r^2$$ > 0,
$$r^2 = (1 + \sqrt{5})/2$$
We have to find cosA/cosC or cosC/cosA
cosA = a/ar$$^2$$ = 1/r$$^2$$
cosC = ar/ar$$^2$$ = 1/r
cosC/cosA = r = $$\sqrt{ (1 + \sqrt{5})/2}$$.
Thus, option B is the right choice.

Let the first term be a and the common difference be d. We have, (a+7d)/(a+16d) = 5/11.

Solving this, we get d=2a. Product of the first and sixth terms is $$a(a+5d) = a(a+10a) = 11a^2 = 176 => a=+- 4$$.
Since the series has only negative terms, a=-4 and d=-8.
Sum of the first 200 terms = (200/2)(2*-4 + 199*- 8) = -160000

given, 2b = a + c...(1)
also given, $$\frac{1}{b^2} - \frac{1}{a^2} = \frac{1}{c^2} - \frac{1}{b^2}$$
$$\frac{(a + b)(a - b)}{a^2b^2} = \frac{(b + c)(b - c)}{c^2b^2}$$
(b - a) = (c - b) as a,b,c are in A.P
so $$\frac{(a + b)}{a^2b^2} = \frac{(b + c)}{c^2b^2}$$
$$ac^2 + bc^2 = a^2b + a^2c$$
$$(ac^2 - a^2c) = (bc^2 - a^2b) = 0$$
ac(c - a) + b(c + a)(c - a) = (c-a)(ac + b(c + a)) = 0
=> c=a or ac + b(c + a) = 0 but c + a = 2b
thus a=b=c or $$ac + 2b^2$$ = 0

Since $$a\ne\ b\ne\ c$$ , $$ac + 2b^2$$ = 0

=> a,b,-c/2 are in GP.

This is a geometric progression with first term equal to 4 and the common ratio equal to 3.
Hence, the sum of the first 10 terms is $$4*\frac{3^{10}-1}{3-1}= 118096$$

2x + y = 28 => ($$\frac{x}{2}$$) + ($$\frac{x}{2}$$) + ($$\frac{x}{2}$$) + ($$\frac{x}{2}$$) + ($$\frac{y}{3}$$) + ($$\frac{y}{3}$$) + ($$\frac{y}{3}$$) = 28

As AM>=GM
=> $$\frac{(\frac{x}{2}) + (\frac{x}{2}) + (\frac{x}{2}) + (\frac{x}{2}) + (\frac{y}{3}) + (\frac{y}{3}) + (\frac{y}{3})}{7} \geq ((\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{y}{3})(\frac{y}{3})(\frac{y}{3})^{\frac{1}{7})}$$
=> $$\frac{28}{7} \geq ((\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{y}{3})(\frac{y}{3})(\frac{y}{3})^{\frac{1}{7})}$$
=> $$(\frac{x^4y^3}{2^4*3^3})^{\frac{1}{7}} \leq 4$$
=> $$\frac{x^4y^3}{2^4*3^3} \leq 2^{14}$$
=> $$x^4y^3 \leq 2^{18}*3^3$$

Let the first term of the AP be a and the common difference be d.
The sixth term of the series will be a+5d
Given that d should be a factor of a+5d
=> a+5d is divisible by d
=> a should be divisible by d
So the required cases are
d = 1, a = 1,2,3.......20
d= 2 , a = 2,4,6.......14
d = 3, a = 3, 6,9
d= 4, a = 4
So the required number of AP’s are 20+7+3+1 = 31

If we consider the number carefully, then nth layer contains n(n+1)/2
$$\sum \frac{n(n+1)}{2} = 8436$$
$$\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}= 8436$$
n(n+1)(n+2) = 50616
n = 36

S = 1 + 2 + 5 + 10 + 17 + 26 +....+ tn

S-S = 1 + (2-1) + (5-2) + (10-5) + (17-10) +..... +[tn-t(n-1)]-tn
0 = 1 + (1 + 3 + 5 + 7 +......+ (n-1) terms) - tn

=> tn = 1 + (n-1)/2 (2 + (n-2)2) = 1 + (n-1)( 1 + n-2) = 1 + $$(n-1)^2$$
=>tn = $$(n^2 - 2n + 2)$$
Sum = (n)(n+1)(2n+1)/6 - n(n+1) + 2n = 50*51*101/6 - 50*51 + 100 = 40475

Let the two numbers be a and b
So, a*b = 144 and a+b = 40
So, the harmonic mean = $$\frac{2*a*b}{a+b}=\frac{288}{40}=7.2$$
 

The series is a geometric progression with first term equal to 19683 and the common ratio equal to $$\frac{1}{3}$$
Hence, the sum of the first six terms equals $$19683*\frac{1-(\frac{1}{3})^6}{1-\frac{1}{3}}=29484$$

This is an arithmetic series with first term equal to 82 and common difference equal to 31
Let the number of digits be equal to n.
So,  82 + (n-1) 31 = 609
or, n = 18
So, the sum equals $$\frac{18}{2}*(82+609)=9*691=6219$$
 

This is a geometric series with first term 8000 and common ratio equal to 0.8
So, the sum of the series is $$\frac{8000}{1-0.8}= 40000$$

The sum of the first 99991 odd numbers is $${99991}^2$$=9998200081

The harmonic mean of 8 and 56 equals $$\frac{2}{\frac{1}{8}+\frac{1}{56}}=2*7=14$$

The arithmetic mean is $$\frac{842321 + 456661}{2}= 649491$$