For doing a certain job, boy is offered ? 5 on the first day, ? 15 on the second day, ? 45 on the third day and so no. How much will the boy earn at the end of seven days?

The number of bacteria at any given time forms a GP whose terms are given by 50, 100, 200, ... where
a = 50, r = 100/50 = 2
T_n = ar^{n -1}
? T₁₂ = 50(2)^{12 - 1}
= 50 x (2)¹¹
= 102400
So, the number of bacteria born in 12th hour is 102400.

As, S₁₂ = S₁₈
So, S₁₁ = S₁₉
S₁₀ = S₂₀
S₉ = S₁₉
.........
.........
.........
S₀ = S₃₀
As, S₀ = 0
So, S₃₀ = 0

As, n is odd.
? Sum of even number = [(n - 1)/2 ] [ (n + 1 )/2]
= 280/2 x 282/2 = 19740

Sum of even numbers = n/2[(n/2) + 1)] = (1672/2) x [(1672/2) + 1]
= 699732

S_n = 120, a = -20, d = 4
S_n = n/2[2a + (n - 1)d ]
? 120 = n/2 [2 x (-20) + (n - 1) x 4]
? 120 = -20n + 2(n -1)n
? 120 = -20n + 2n² - 2n
? 120 = 2n² - 22n
? 2n² - 22n - 120 = 0
? n²-11n - 60 = 0
? n² - 15n + 4n - 60 = 0
? n(n - 15) + 4(n - 15) = 0
? (n + 4) (n - 15) = 0
? n = -4 or 15
? n = 15

Given, F = 2E + 4Y ...(I)
?EY = 4?3
? EY = 48 ...(II)
and 2EY/ E + Y = 6 ? E + Y = 16 ...(III)
Now, (E -Y)² = (E + Y)² - 4EY
= (16)² - 4 x 48
= 256 - 192 = 64
? E - Y = 8 ...(iv)
From Eqs. (iii) and (iv), we get
E = 12 and Y = 4
From Eq. (i) F = 2 x 12 + 4 x 4 = 40 yr

The nth term of a GP is ar^{n - 1} ,
5th term = ar^{5 - 1} = ar⁴ = 81
1st term = a = 16
? r⁴ = 81/16
? r = ?81/16 = 3/2
? 4th term = ar^{4 - 1} = ar³ = 16 x 3/2 x 3/2 x 3/2 = 54

Let term = l = ar^{n - 1} a = 5 and l = 20480
r = 20/5 = 4
? 20480 = 5 x (4)^n-1
(4)^n-1 = 20480/5 = 4096 = (4)⁶
n - 1 = 6
? n = 7

(a + b)/2 = 25
a + b = 50
?ab = 7
ab = 49
Hence, A can either be 7 or 49.
So, 49 is the answer.

The second and fourth term of the HP are 1/6 and 1/14 respectively,
Hence, for the corresponding AP, the second term is 6 and fourth term is 14.
Hence a + d = 6 and a + 3d = 14
? 2d = 8
? d = 4 and a = 2

Hence, the 10th term of this AP = a + (10 -1)d
= 2 + (10 - 1)x 4 = 2 + 9 x 4
= 2 + 36 = 38
Hence, for the corresponding HP, the 10th term is 1/38.