nth term from sum of n terms A.P.

1. Concept Overview

Sometimes, instead of the general term [Tₙ], we are given the sum of first n terms, i.e. [Sₙ].
From this information, we are asked to find the nᵗʰ term of the A.P.

This concept is extremely useful when:

• [Sₙ] is given directly as a formula
• Finding the actual A.P. is difficult
• Questions involve comparison of terms

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2. Key Idea and Formula

The fundamental relationship is:

[\boxed{T_n = S_n − S_{n-1}}]

Reason:
[Sₙ] is the sum of first [n] terms
[Sₙ₋₁] is the sum of first [n−1] terms
So, subtracting them leaves only the [nᵗʰ] term.

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3. Step-by-Step Method

To find [Tₙ] from [Sₙ]:

1. Write the given expression for [Sₙ]
2. Replace [n] by [(n−1)] to get [Sₙ₋₁]
3. Subtract:
   [Tₙ = Sₙ − Sₙ₋₁]
4. Simplify carefully

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4. Examples with Solutions

Example 1

If [Sₙ = 3n² + 5n], find the nᵗʰ term.

Solution:
[Sₙ = 3n² + 5n]

[Sₙ₋₁ = 3(n−1)² + 5(n−1)]
[Sₙ₋₁ = 3(n² − 2n + 1) + 5n − 5]
[Sₙ₋₁ = 3n² − 6n + 3 + 5n − 5]
[Sₙ₋₁ = 3n² − n − 2]

[Tₙ = Sₙ − Sₙ₋₁]
[Tₙ = (3n² + 5n) − (3n² − n − 2)]
[Tₙ = 6n + 2]

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Example 2

If [Sₙ = n(2n + 1)], find [Tₙ].

Solution:
[Sₙ = 2n² + n]

[Sₙ₋₁ = (n−1)(2n−1)]
[Sₙ₋₁ = 2n² − 3n + 1]

[Tₙ = (2n² + n) − (2n² − 3n + 1)]
[Tₙ = 4n − 1]

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Example 3

If [Sₙ = n² + 3n], find the 10ᵗʰ term.

Solution:
First find [Tₙ]:

[Sₙ₋₁ = (n−1)² + 3(n−1)]
[Sₙ₋₁ = n² + n − 2]

[Tₙ = (n² + 3n) − (n² + n − 2)]
[Tₙ = 2n + 2]

[T₁₀ = 2(10) + 2 = 22]

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Example 4

If [Sₙ = 5n² − n], find [Tₙ].

Solution:
[Sₙ₋₁ = 5(n−1)² − (n−1)]
[Sₙ₋₁ = 5n² − 11n + 6]

[Tₙ = (5n² − n) − (5n² − 11n + 6)]
[Tₙ = 10n − 6]

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Example 5

If [Sₙ = n(3n − 2)], find the A.P.

Solution:
[Sₙ = 3n² − 2n]

[Sₙ₋₁ = 3n² − 8n + 5]

[Tₙ = 6n − 5]

Comparing with [Tₙ = a + (n−1)d]:

[d = 6]
[a = 1]

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5. Important Observations

• If [Sₙ] is a quadratic expression, the sequence is an A.P.
• [Tₙ] will always be a linear expression
• This method avoids finding [a] and [d] separately

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6. Conceptual Questions with Solutions

1. Why do we subtract Sₙ₋₁ from Sₙ to get Tₙ?

[Sₙ] contains the sum of first [n] terms, while [Sₙ₋₁] contains the sum of first [n−1] terms. Subtracting removes all previous terms and leaves only the nᵗʰ term.

2. Can this method work if Sₙ is not quadratic?

No. If [Sₙ] is not a quadratic polynomial, the resulting [Tₙ] will not be linear, and the sequence will not be an A.P.

3. Why must Tₙ be linear for an A.P.?

In an A.P., the difference between consecutive terms is constant. This is only possible when [Tₙ] is a linear function of n.

4. Can we find a and d directly from Sₙ?

Yes. After finding [Tₙ], compare it with [a + (n−1)d] to obtain a and d.

5. Is this method faster than using sum formula?

Yes. It avoids unnecessary algebra and gives [Tₙ] directly using subtraction.

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7. FAQs / Common Misconceptions

1. Can I directly differentiate Sₙ to get Tₙ?

No. Differentiation is not valid here. The correct method is subtraction, not calculus.

2. Is S₁ always equal to a?

Yes. Since [S₁] contains only the first term, it is equal to a.

3. What happens if Sₙ is cubic?

The sequence will not be an A.P. because [Tₙ] will not be linear.

4. Can this method be used in reverse?

Yes. If [Tₙ] is known, [Sₙ] can be obtained by summation.

5. Do we always need to simplify fully?

Yes. Incomplete simplification often leads to wrong [a] and [d].

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8. Practice Questions with Step-by-Step Solutions

Question 1. If [Sₙ = 4n² + 3n], find [Tₙ].

Step-by-Step Solution:
[Sₙ₋₁ = 4(n−1)² + 3(n−1)]
[Sₙ₋₁ = 4n² − 5n + 1]

[Tₙ = (4n² + 3n) − (4n² − 5n + 1)]
[Tₙ = 8n − 1]

Question 2. If [Sₙ = n(5n − 2)], find the A.P.

Solution:
[Sₙ = 5n² − 2n]
[Sₙ₋₁ = 5n² − 12n + 7]

[Tₙ = 10n − 9]
[d = 10], [a = 1]

Question 3. If [Sₙ = 2n² − n], find the 15ᵗʰ term.

Solution:
[Tₙ = 4n − 3]
[T₁₅ = 57]

Question 4. If [Sₙ = n(n + 1)], find [Tₙ].

Solution:
[Tₙ = 2n]

Question 5. If [Sₙ = 6n² − 5n], find the first term.

Solution:
[Tₙ = 12n − 11]
[a = 1]