Sum of First n Terms (Derivation) MCQs Quiz | Class 10

Understanding the Sum of First ‘n’ Terms of an AP

An Arithmetic Progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by ‘d’. The sum of the first ‘n’ terms of an AP is a fundamental concept in mathematics, and its derivation reveals an elegant method for calculating this sum efficiently.

Key Formulas:

• The nth term of an AP: a_n = a + (n-1)d, where ‘a’ is the first term, ‘n’ is the number of terms, and ‘d’ is the common difference.
• The sum of the first ‘n’ terms of an AP: S_n = n/2 [2a + (n-1)d]
• An alternative formula for the sum of the first ‘n’ terms, if the last term ‘l’ is known: S_n = n/2 [a + l]

Derivation Logic of S_n = n/2 [2a + (n-1)d]

The derivation of the sum formula is a classic example of mathematical elegance, often attributed to the young Carl Friedrich Gauss. Here’s a step-by-step breakdown:

Step 1: Write the sum in forward order.

Let S_n be the sum of the first ‘n’ terms of an AP. The terms are a, a+d, a+2d, …, a+(n-1)d.

S_n = a + (a+d) + (a+2d) + … + (a+(n-2)d) + (a+(n-1)d) — (Equation 1)

Step 2: Write the sum in reverse order.

Now, write the same sum, but with the terms in reverse order. The last term is l = a+(n-1)d, the second to last is l-d = a+(n-2)d, and so on.

S_n = (a+(n-1)d) + (a+(n-2)d) + … + (a+d) + a — (Equation 2)

Step 3: Add Equation 1 and Equation 2.

When you add the corresponding terms from Equation 1 and Equation 2, a remarkable pattern emerges:

2S_n = [a + (a+(n-1)d)] + [(a+d) + (a+(n-2)d)] + … + [(a+(n-1)d) + a]

Notice that each pair of terms sums up to the same value: (2a + (n-1)d).

• First pair: a + (a+(n-1)d) = 2a + (n-1)d
• Second pair: (a+d) + (a+(n-2)d) = a+d+a+nd-2d = 2a + (n-1)d
• …and so on, for all ‘n’ terms.

Since there are ‘n’ such pairs, each summing to (2a + (n-1)d), we get:

2S_n = n * [2a + (n-1)d]

Step 4: Solve for S_n.

Divide both sides by 2:

S_n = n/2 [2a + (n-1)d]

Connecting to S_n = n/2 [a + l]

We know that the last term, l, is given by l = a + (n-1)d. We can substitute this into the derived formula:

S_n = n/2 [a + {a + (n-1)d}]

S_n = n/2 [a + l]

Both formulas are equally valid and useful depending on the information available.

Quick Revision:

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant (d).
• First term (a): The starting term of the AP.
• Common difference (d): The constant difference between any two consecutive terms.
• Number of terms (n): How many terms are being summed.
• nth term (a_n or l): The term at position ‘n’. Formula: a_n = a + (n-1)d.
• Sum of ‘n’ terms (S_n): The total of all terms up to the nth term.
• Derivation essence: Write the sum twice (forward and reverse), add them, and exploit the constant sum of paired terms.

Practice Questions (without options):

1. Find the sum of the first 20 terms of an AP whose first term is 5 and the common difference is 3.
2. An AP has 12 terms. If the first term is -2 and the last term is 20, find the sum of all terms.
3. The sum of the first 15 terms of an AP is 300. If the first term is 5, find the common difference.
4. How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
5. Derive the formula for the sum of the first ‘n’ terms of an AP when the first term ‘a’ and the last term ‘l’ are given.