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An arithmetic progression (AP) is a sequence of numbers in which each term is obtained by adding a fixed number to the preceding term. One important concept in AP is the sum of the first n terms in the sequence, which can be calculated using a simple formula. This formula is a powerful tool for solving problems involving arithmetic sequences, and it has applications in various fields such as finance, physics, and mathematics.
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Key Terms: Arithmetic Progression, Terms, Sum, Common Difference, Sequence, Series
Sum of First N Terms of an AP Formula
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Arithmetic Progression, commonly abbreviated as AP, is a type of sequence in mathematics in which the difference between any two consecutive terms is constant. This constant difference is known as the common difference and is denoted by the letter d. The sum of the first n terms in an AP is given by the formula:
Sn = n/2 [2a + ( n – 1) d]
- Where Sn represents the sum of the first n terms,
- a represents the first term of the sequence,
- d represents the common difference,
- n represents the number of terms in the sequence.
Sum of n terms of an A. P.
Sn = \(\frac{n}{2}\) {2a + ( n – 1) d}
The first term (a) represents the starting point of the sequence. For example, if we have an AP with a first term of 2 and a common difference of 3, the sequence would look like this: 2, 5, 8, 11, 14, and so on.
The common difference (d) represents the amount by which each term in the sequence differs from the previous term.
Using the same example as above, the common difference is 3 since each term increases by 3.
The number of terms (n) represents how many terms there are in the sequence. For example, if we have an AP with a first term of 2, and a common difference of 3, and we want to find the sum of the first 5 terms, n would be equal to 5.
Example: Find the sum of the first 10 terms in the sequence 3, 7, 11, 15, …
Solution: The first term, a = 3
The common difference, d = 7 - 3 = 4
The number of terms, n = 10
Using the formula, we have:
Sn = n/2 [2a + (n-1)d]
S10 = 10/2 [2(3) + (10-1)(4)]
S10 = 5 [6 + 36]
S10 = 5 x 42
S10 = 210
Therefore, the sum of the first 10 terms in the sequence is 210.
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Arithmetic Sequence | Infinite Series Formula | Arithmetic Sequence Explicit Formula |
Series Formula | Difference between Sequence and Series | Arithmetic Sequence Recursive Formula |
Formula Derivation
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Let a be the first term and d be the common difference of an AP. Then the nth term of the AP is given by:
an = a + (n-1)d
To find the sum of the first n terms of the AP, we can add the terms together:
Sn = a + (a+d) + (a+2d) + ... + [a + (n-2)d] + [a + (n-1)d]
Re-arranging the terms, we can write:
Sn = (a + a + (n-1)d) + (a+d + a+(n-2)d) + ... + [(a+(n-2)d) + (a+(n-1)d)]
Simplifying, we get:
Sn = n/2 [2a + (n-1)d]
Formula Breakdown
Now let's examine the formula itself. The formula for the sum of the first n terms in an AP is:
Sn = n/2 [2a + (n-1)d]
This formula can be broken down into two parts:
The first part, n/2, represents the average of the first and last terms in the sequence. This is because the sum of the first and last terms is equal to 2a + (n-1)d, and the average of these two terms is (2a + (n-1)d)/2, which simplifies to a + (n-1)d/2. Multiplying this by n/2 gives us the sum of the first n terms in the sequence.
The second part, 2a + (n-1)d, represents the sum of the first and last terms in the sequence. We multiply this by n/2 to get the sum of the first n terms.
Finding Individual Terms Using Sum Formula
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The formula to find the sum of the first n terms of an arithmetic progression (AP) is:
Sn = n/2[2a + (n-1)d]
where a is the first term, d is a common difference, and n is the number of terms.
To find an individual term in an AP from the sum formula, we can use the following steps:
- Rearrange the sum formula to isolate the term we want to find.
For example, if we want to find the kth term of the AP, we can write:
ak = [Sn - (n-k)d]/2
- Plug in the values for Sn, n, and d from the problem. If the problem gives us the values of a and d, we can use the formula for Sn to find n first, and then plug in the values.
- Solve for ak using basic algebra.
Example: Find the 5th term of an AP with a first term of 2 and a common difference of 3, given that the sum of the first 10 terms is 145.
Solution: We can use the sum formula to find n:
Sn = n/2[2a + (n-1)d]
145 = 10/2[2(2) + (10-1)3]
145 = 5[4 + 27]
145 = 155
n = 10
Now we can plug in the values for a, d, n, and k into the formula we derived earlier:
ak = [S_n - (n-k)d]/2
a5 = [145 - (10-5)3]/2
a5 = [145 - 15]/2
a5 = 65/2
a5 = 32.5
Therefore, the 5th term of the AP is 32.5.
Sum of AP Formula for an Infinite AP
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The sum of an infinite arithmetic progression (AP) can be calculated using the following formula:
Sum = (a) / (1 - d)
where the a is the initial value of the progression and d is the common difference.
- Note that this formula is valid only when the common difference is less than 1 in absolute value.
- If the common difference is greater than 1 in absolute value, then the series diverges and does not have a finite sum.
Also, it's important to note that in practice, we cannot actually add up an infinite number of terms, so we use the formula to find an approximate value of the sum based on a finite number of terms.
Sum of Square Series
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The sum of squares of the first n natural numbers can be calculated using the following formula:
Sum = n(n+1)(2n+1)/6
where n is the number of terms in the series.
Example: Find the sum of squares of the first 5 natural numbers
Solution: Sum = 5(5+1)(2(5)+1)/6 = 5(6)(11)/6 = 55
So the sum of squares of the first 5 natural numbers is 55.
Sum of Cubic Series
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The sum of cubes of the first n natural numbers can be calculated using the following formula:
Sum = [n(n+1)/2]2
where n is the number of terms in the series.
Example: Find the sum of cubes of the first 4 natural numbers
Solution: Sum = [4(4+1)/2]2 = 102 = 100
So the sum of cubes of the first 4 natural numbers is 100.
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Arithmetic Mean Formula | Algebra Formula | Algebraic Expressions and Identities |
Important General Formulas
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Here are some of the top general formulas used in finding the sum of n terms of an arithmetic progression (AP):
Sum of first n terms | n/2 [2a + (n – 1)d] |
nth term | a + (n – 1)d |
Sum of natural numbers | n (n + 1) / 2 |
Sum of first n even terms | n(n – 1) |
Sum of first n odd terms | 2n |
Sum of squares of first n terms | [n (n + 1) (2n + 1)] / 6 |
Sum of cubes of first n terms | [n/2 (n + 1)]2 |
Things to Remember
- The formula for the sum of n terms of an AP is Sn = n/2 [2a + (n-1)d], where a is the first term, d is the common difference, and n is the number of terms.
- If the nth term of the AP is not given, it can be found by an = a + (n-1)d.
- The sum of the first n even or odd terms of an AP can be found using the same formula as the sum of the first n terms, but with a and d adjusted to the first even or odd term and the common difference between consecutive even or odd terms, respectively.
- If the AP is a finite series (i.e., it has a last term), you can also use the formula Sn = (n/2)(a1 + an) to find the last term of the series. Rearranging the formula gives an = 2Sn/n - a1.
Sample Questions
Ques. Find the sum of the first 8 terms of the AP 5, 10, 15, … (1 mark)
Ans: Here, a = 5 and d = 5. Using the formula for the sum of n terms, we get S8 = 8/2 [2(5) + (8-1)5] = 240. Therefore, the sum of the first 8 terms is 240.
Ques. Find the sum of the first 10 terms of the AP -4, -1, 2, … (1 mark)
Ans: Here, a = -4 and d = 3. Using the formula for the sum of n terms, we get S10 = 10/2 [2(-4) + (10-1)3] = 45. Therefore, the sum of the first 10 terms is 45.
Ques. If the sum of squares of the first n natural numbers is 385, find the value of n. (3 marks)
Ans: Using the formula for the sum of squares and setting it equal to 385, we have:
n(n+1)(2n+1)/6 = 385
Multiplying both sides by 6 and simplifying, we get:
n3 + n2 - 77n - 385 = 0
Using trial and error or a computer program, we can find that n=10 is a solution to this equation. Therefore, the value of n is 10.
Ques. Find the sum of the first 5 terms of the AP -3, -1, 1, … (2 marks)
Ans: Here, a = -3 and d = 2. Using the formula for the sum of n terms, we get S5 = 5/2 [2(-3) + (5-1)2] = – 5. Therefore, the sum of the first 5 terms is – 5.
Ques. Find the sum of the first 6 terms of the AP 1/2, 1, 3/2, … (2 marks)
Ans: Here, a = 1/2 and d = 1/2. Using the formula for the sum of n terms, we get S6 = 6/2 [2(1/2) + (6-1)(1/2)] = 9/2. Therefore, the sum of the first 6 terms is 9/2.
Ques. If the sum of cubes of the first n natural numbers is 36, find the value of n. (3 marks)
Ans: Using the formula for the sum of cubes and setting it equal to 36, we have:
[n(n+1)/2]2 = 36
Taking the square root of both sides and simplifying, we get:
n(n+1)/2 = 6
Multiplying both sides by 2 and rearranging, we get:
n2 + n - 12 = 0
Using factoring or the quadratic formula, we can find that n = 3 or n = -4 are solutions to this equation. Since we are looking for a positive value of n, the value of n is 3.
Ques. Find the sum of squares of the first 20 even natural numbers. (3 marks)
Ans: We can rewrite the sum of squares of the first n natural numbers as the sum of squares of the first n/2 even natural numbers. Therefore, we have:
Sum = (n/2)((n/2)+1)(2(n/2)+1)/6
Setting n=20, we get:
Sum = 10(11)(21)/6
= 385
So the sum of squares of the first 20 even natural numbers is 385.
Ques. Find the sum of the first 15 terms of the AP 1, -1/2, -2, … (2 marks)
Ans: Here, a = 1 and d = -3/2. Using the formula for the sum of n terms, we get S15 = 15/2 [2(1) + (15-1)(-3/2)] = -120. Therefore, the sum of the first 15 terms is – 120.
Ques. Find the sum of the first 9 terms of the AP -6, -1, 4, … (2 marks)
Ans. Here, a = – 6 and d = 5. Using the formula for the sum of n terms, we get S9 = 9/2 [2(-6) + (9-1)(5)] = 135. Therefore, the sum of the first 9 terms is 135.
Ques. Find the sum of cubes of the first 10 even natural numbers. (3 marks)
Ans. We can rewrite the sum of cubes of the first n natural numbers as the sum of cubes of the first n/2 even natural numbers. Therefore, we have:
Sum = [(n/2)(n+1)]2
= (5*6)2
= 900
So the sum of cubes of the first 10 even natural numbers is 900.
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