Sequences and Series

x = Σ a? = 1/ (1-a); y = Σ b? = 1/ (1-b); z = Σ c? = 1/ (1-c)

Now,
a, b, c → AP
1-a, 1-b, 1-c → AP
1/ (1-a), 1/ (1-b), 1/ (1-c) → HP
x, y, z → HP

. Let the terms in Arithmetic Progression be a – 2d, a – d, a, a + d, a + 2d.
Sum of terms: (a – 2d) + (a – d) + a + (a + d) + (a + 2d) = 5a.
5a = 25 ⇒ a = 5.
Product of terms: (5 – 2d) (5 – d) (5) (5 + d) (5 + 2d) = 2520.
5 (25 – 4d²) (25 – d²) = 2520.
(25 – 4d²) (25 – d²) = 504.
625 – 25d² – 100d² + 4d? = 504.
4d? – 125d² + 121 = 0.
Factoring the equation: (4d² - 121) (d² - 1) = 0.
So, d² = 1 or d² = 121/4.
d = ±1 or d = ±11/2.
If d = ±1, the terms are 3, 4, 5, 6, 7.
If d = ±11/2, the terms are -6, -1/2, 5, 21/2, 16.
The largest term is 5 + 2d = 5 + 2 (11/2) = 5 + 11 = 16.

1/16, a, b are in GP. So, a² = b/16 .
Also, a, b, 1/6 are in AP. So, 2b = a + 1/6.
From the first equation, b = 16a².
Substitute into the second: 2 (16a²) = a + 1/6 => 32a² - a - 1/6 = 0.
192a² - 6a - 1 = 0.
The solution appears to solve a different problem.

S? (x) = log? ¹? ²x + log? ¹? ³x + .
This is incorrect; the bases are numbers, not powers of 'a'. Let's assume the bases are 1/2, 1/3, 1/6, 1/11, .
The series is S' = 2, 3, 6, 11, 18, .
The differences are 1, 3, 5, 7, . which is an AP.
The n-th term t? is a quadratic in n.
t? = An² + Bn + C.
t? =A+B+C=2
t? =4A+2B+C=3
t? =9A+3B+C=6
Solving these, we get 3A+B=1 and 5A+B=3, which gives 2A=2, A=1. Then B=-2, C=3.
t? = n² - 2n + 3 = (n-1)² + 2.
The solution confirms this finding t? = 2 + (n-1)².

The problem provides non-standard formulas for sums S? and S? :
S? = n [a + (2n-1)d]
S? = 2n [a + (4n-1)d]
We are given S? - S? = 1000.
S? - S? = 2n [a + (4n-1)d] - n [a + (2n-1)d]
= (2na + 8n²d - 2nd) - (na + 2n²d - nd)
= na + 6n²d - and = n [a + (6n-1)d].
So, n [a + (6n-1)d] = 1000.
We need to find S? n. Assuming the pattern S? n = kn [a + (2kn-1)d], then S? n would be 3n [a + (6n-1)d].
S? n = 3 * (n [a + (6n-1)d]) = 3 * 1000 = 3000.

To find the sum Σ[r=0 to 6] (?C?)².
This is the coefficient of x? in the expansion of (1+x)?(x+1)? = (1+x)¹².
By the binomial theorem, (1+x)¹² = Σ[k=0 to 12] ¹²C? x?.
The coefficient of x? is ¹²C?.
¹²C? = (12 * 11 * 10 * 9 * 8 * 7) / (6 * 5 * 4 * 3 * 2 * 1) = 11 * 2 * 3 * 2 * 7 = 924.

We need to find the remainder of (2021)³? ² when divided by 17.
First, find the remainder of 2021 divided by 17.
2021 = 17 * 118 + 15.
So, 2021 ≡ 15 (mod 17).
Also, 15 ≡ -2 (mod 17).
So, (2021)³? ² ≡ (-2)³? ² (mod 17).
(-2)³? ² = 2³? ² = (2? )? ⋅ 2² = 16? ⋅ 4.
Since 16 ≡ -1 (mod 17),
16? ⋅ 4 ≡ (-1)? ⋅ 4 (mod 17).
≡ 1 ⋅ 4 (mod 17)
≡ 4 (mod 17).
The remainder is 4.