✨ Complete GP Adventure ✨

For sequence (i):

First term (a) = 5

Common ratio (r) = -3/5

Since |r| = 3/5 < 1, we use:

Sₙ = 5(1 - (-3/5)ⁿ)/(1 - (-3/5)) = (25/8)(1 - (-3/5)ⁿ)

For sequence (ii):

First term (a) = 256

Common ratio (r) = 1/4

Since |r| = 1/4 < 1, we use:

Sₙ = 256(1 - (1/4)ⁿ)/(3/4) = (1024/3)(1 - (1/4)ⁿ)

First term (a) = 5

Common ratio (r) = 3

Since r > 1, we use:

S₆ = 5(3⁶ - 1)/(3 - 1) = 5(729 - 1)/2 = 1820

Given: r = 5, S₆ = 46872, n = 6

Using sum formula for r > 1:

46872 = a(5⁶ - 1)/4

46872 = a(15624)/4

a = 46872 / 3906 = 12

For sequence (i):

a = 9, r = 3/9 = 1/3

Since |r| < 1, sum to infinity exists:

S∞ = 9/(1 - 1/3) = 9/(2/3) = 27/2

For sequence (ii):

a = 21, r = 14/21 = 2/3

Since |r| < 1, sum to infinity exists:

S∞ = 21/(1 - 2/3) = 21/(1/3) = 63

Given: a = 8, S∞ = 32/3

32/3 = 8/(1 - r)

32(1 - r) = 24

1 - r = 24/32 = 3/4

r = 1 - 3/4 = 1/4

For series (i):

This can be written as:

S = 4(0.1 + 0.11 + 0.111 + ...)

Each term is 1/9(1 - 1/10ⁿ), so:

Sₙ = (4/9)[n - (10/9)(1 - (1/10)ⁿ)]

For series (ii):

This can be written as:

S = 3(1 + 11 + 111 + ...)

Each term is (10ⁿ - 1)/9, so:

Sₙ = (1/3)[10(10ⁿ - 1)/9 - n]

First term (a) = 3

Common ratio (r) = 2

Last term (l) = 1536

Find number of terms (n):

1536 = 3 × 2ⁿ⁻¹ ⇒ 2ⁿ⁻¹ = 512 ⇒ n-1 = 9 ⇒ n = 10

S₁₀ = 3(2¹⁰ - 1)/(2 - 1) = 3(1024 - 1) = 3069

This forms a GP where:

1st set: 4 letters

2nd set: 4 × 4 = 16 letters

...

8th set: 4⁸ letters

Total letters = 4 + 16 + ... + 4⁸

This is a GP with a = 4, r = 4, n = 8

S₈ = 4(4⁸ - 1)/3 = 4(65536 - 1)/3 = 87380

Total cost = 87380 × ₹2 = ₹174,760

Let x = 0.123123123...

Then 1000x = 123.123123...

Subtract: 999x = 123

x = 123/999 = 41/333

Each term in the series is a finite geometric series:

kth term = xᵏ + xᵏ⁻¹y + ... + yᵏ = (xᵏ⁺¹ - yᵏ⁺¹)/(x - y)

Thus, S = Σ[(xᵏ⁺¹ - yᵏ⁺¹)/(x - y)] from k=1 to n

(x - y)S = Σ(xᵏ⁺¹ - yᵏ⁺¹) from k=1 to n

= (x² + x³ + ... + xⁿ⁺¹) - (y² + y³ + ... + yⁿ⁺¹)

= x²(xⁿ - 1)/(x - 1) - y²(yⁿ - 1)/(y - 1)

But the given RHS is different, suggesting possible typo in original problem.

The correct version should likely have x² and y² in numerators.