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Kumar Rohan

Physics and Mathematics

Example 4 Sum of A.P.

Practice Questions with Step-by-Step Solutions

Question 1. The sum of n terms of two A.P.s are in the ratio [ (2n + 1) : (5n + 3) ]. Find the ratio of their 10ᵗʰ terms.

Step-by-Step Solution:

Let the two A.P.s have first terms [a₁, a₂] and common differences [d₁, d₂].

Given:
[\dfrac{Sₙ₁}{Sₙ₂} = \dfrac{2n + 1}{5n + 3}]

For large [n], the ratio of sums depends mainly on the coefficient of [n²].

Since [Sₙ = \dfrac{n}{2} (2a + (n−1)d)], the coefficient of [n²] is proportional to [d].

Hence,
[\dfrac{d₁}{d₂} = \dfrac{2}{5}]

The 10ᵗʰ terms are:
[T₁₀ = a₁ + 9d₁], [T₁₀’ = a₂ + 9d₂]

Ratio depends mainly on [d₁ : d₂].

Conclusion:
The ratio of their 10ᵗʰ terms is [2 : 5].

Question 2. The sums of n terms of two A.P.s are in the ratio [ (3n + 4) : (6n + 7) ]. Find the ratio of their 12ᵗʰ terms.

Step-by-Step Solution:

Let the first A.P. have first term [a₁] and common difference [d₁].
Let the second A.P. have first term [a₂] and common difference [d₂].

The sum of n terms of an A.P. is
[Sₙ = \dfrac{n}{2} (2a + (n − 1)d)]

For large values of [n], the dominant part of [Sₙ] is proportional to [n²],
and the coefficient of [n²] depends only on the common difference.

Given:
[\dfrac{Sₙ₁}{Sₙ₂} = \dfrac{3n + 4}{6n + 7}]

Compare coefficients of [n]:
[d₁ : d₂ = 3 : 6 = 1 : 2]

The 12ᵗʰ terms are
[T₁₂ = a₁ + 11d₁] and [T₁₂’ = a₂ + 11d₂]

Since the same term number is taken in both A.P.s, the ratio depends on
[d₁ : d₂].

Conclusion:
The ratio of their 12ᵗʰ terms is [1 : 2].

Question 3. The sum of n terms of two A.P.s are in the ratio [ (4n − 1) : (2n + 3) ]. Find the ratio of their 15ᵗʰ terms.

Step-by-Step Solution:

Let the A.P.s be [a₁, d₁] and [a₂, d₂].

Sum of n terms:
[Sₙ = \dfrac{n}{2} (2a + (n − 1)d)]

Given ratio:
[\dfrac{Sₙ₁}{Sₙ₂} = \dfrac{4n − 1}{2n + 3}]

The dominant term for large [n] is proportional to [n²].

Compare coefficients of [n]:
[d₁ : d₂ = 4 : 2 = 2 : 1]

The 15ᵗʰ terms are
[T₁₅ = a₁ + 14d₁] and [T₁₅’ = a₂ + 14d₂]

Ratio depends on [d₁ : d₂].

Conclusion:
The ratio of their 15ᵗʰ terms is [2 : 1].

Question 4. The sum of n terms of two A.P.s are in the ratio [ (5n + 2) : (10n + 9) ]. Find the ratio of their 8ᵗʰ terms.

Step-by-Step Solution:

Let the two A.P.s have common differences [d₁] and [d₂].

Given:
[\dfrac{Sₙ₁}{Sₙ₂} = \dfrac{5n + 2}{10n + 9}]

For large [n], ratio depends on coefficients of [n].

Therefore,
[d₁ : d₂ = 5 : 10 = 1 : 2]

The 8ᵗʰ terms are
[T₈ = a + 7d]

Same position term implies same multiplier of [d].

Conclusion:
The ratio of their 8ᵗʰ terms is [1 : 2].

Question 5. The sums of n terms of two A.P.s are in the ratio [ (7n − 3) : (14n + 1) ]. Find the ratio of their 20ᵗʰ terms.

Step-by-Step Solution:

Use sum formula of A.P.

Given ratio:
[\dfrac{Sₙ₁}{Sₙ₂} = \dfrac{7n − 3}{14n + 1}]

Dominant coefficients give
[d₁ : d₂ = 7 : 14 = 1 : 2]

The 20ᵗʰ terms are
[T₂₀ = a + 19d]

Conclusion:
The ratio of their 20ᵗʰ terms is [1 : 2].

Question 6. The sum of n terms of two A.P.s are in the ratio [ (9n + 5) : (3n + 1) ]. Find the ratio of their 6ᵗʰ terms.

Step-by-Step Solution:

Given ratio of sums.

Compare coefficients of [n]:
[d₁ : d₂ = 9 : 3 = 3 : 1]

The 6ᵗʰ terms are
[T₆ = a + 5d]

Conclusion:
The ratio of their 6ᵗʰ terms is [3 : 1].

Question 7. The sum of n terms of two A.P.s are in the ratio [ (n + 4) : (2n + 1) ]. Find the ratio of their 9ᵗʰ terms.

Step-by-Step Solution:

Given sum ratio.

Dominant coefficients:
[d₁ : d₂ = 1 : 2]

The 9ᵗʰ terms are
[T₉ = a + 8d]

Conclusion:
The ratio of their 9ᵗʰ terms is [1 : 2].

Question 8. The sums of n terms of two A.P.s are in the ratio [ (11n − 2) : (22n + 3) ]. Find the ratio of their 14ᵗʰ terms.

Step-by-Step Solution:

Compare coefficients of [n].

[d₁ : d₂ = 11 : 22 = 1 : 2]

The 14ᵗʰ terms are
[T₁₄ = a + 13d]

Conclusion:
The ratio of their 14ᵗʰ terms is [1 : 2].

Question 9. The sum of n terms of two A.P.s are in the ratio [ (6n + 7) : (3n + 5) ]. Find the ratio of their 5ᵗʰ terms.

Step-by-Step Solution:

Compare dominant coefficients.

[d₁ : d₂ = 6 : 3 = 2 : 1]

The 5ᵗʰ terms are
[T₅ = a + 4d]

Conclusion:
The ratio of their 5ᵗʰ terms is [2 : 1].

Question 10. The sum of n terms of two A.P.s are in the ratio [ (8n + 9) : (4n + 1) ]. Find the ratio of their 12ᵗʰ terms.

Step-by-Step Solution:

Compare coefficients of [n].

[d₁ : d₂ = 8 : 4 = 2 : 1]

The 12ᵗʰ terms are
[T₁₂ = a + 11d]

Conclusion:
The ratio of their 12ᵗʰ terms is [2 : 1].

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