Sum to n terms of arithmatic progression

Math Formulas

Sum to n terms of arithmatic progression

Definition and description of formula :-
Let us consider how the sum of all natural numbers between 1-100 can be derived without a formula.

In this sequence, the sum of numbers can be represented as such:

1. Sum = 1+2+3+4+5+6….+97+98+99+100
Even when the order is reversed, the sum does not change:

2. Sum = 100+99+98+97….6+5+4+3+2+1
Hence, when both equations 1 and 2 are added, we get:

2 x sum = (100+1) + (99+2) + (98+3) + (97+4) + … (4+97) + (3+98) + (2+99) + (1+100)
2 x sum = 101+ 101 + 101 + 101 + … (4+97) + (3+98) + (2+99) + (1+100)
2 x sum = 101 (100 terms)
2sum = 101(100)

Sum to n terms of arithmatic progression
sum = 5,050

This method can be used to calculate the sum of natural numbers like 1000, 10,000 or even 100,000. Hence, knowing the last term in the sequence, this method can be used to derive the formula needed to figure out the nth term in any given sequence.

If we express the first term in the academic progression as a, the common difference between each consecutive term as d, and the last term as l.

1. Sum = a + (a + d) + (a + 2d) + (a + 3d)…+ (l – 3d) + (l – 2d) + (l – d) + l,Where l = a + (n – 1)d
In reverse order, the sum remains the same:

2. Sum = l + (l – d) + (l – 2d) + (l – 3d) + … (a + 3d) + (a + 2d) + (a + d) + a
Adding equations a and b, we get:
2 x sum = (a + l) + [(a +d) + (l – d)] … + [(l – d) + (a + d)] + (l + a)]
• 2 x sum = (a + l) + (a + l)…+ (a + l) + (a + l)
• 2 x sum = n x (a + l)

Sum to n terms of arithmatic progression
Substituting l with the previous equation above (where l = a + (n – 1)d): Sum to n terms of arithmatic progression

Example 1 :-Find the sum of first 500 natural numbers.

Solution :-
Let s be the required sum.
Therefore, s = 1 + 2 + 3 + 4 + 5 + .................... + 500
Clearly, it is an arithmetic progression whose first term = 1, last term = 500 and number of terms = 500.

Therefore, s = 50025002(500 + 1), [using the formula s = n2n2(a + l)]
= 225(501)
= 112725

Therefore, the sum of first 100 natural numbers is 112725.

Example 2 :-Find the sum of the series: 7 + 15 + 23 + 31 + 39 + 47 + ……….. + 255

Solution :-
First term of the given arithmetic series = 7
Second term of the given arithmetic series = 15
Third term of the given arithmetic series = 23
Fourth term of the given arithmetic series = 31
Fifth term of the given arithmetic series = 39
Now, second term - first term = 15 - 7 = 8
Third term - second term = 23 - 15 = 8
Fourth term - third term = 31 - 23 = 8

Therefore, the given sequence is ann arithmetic series with the common difference 8.
Let there be n terms in the given arithmetic series. Then
Ann = 255
⇒ a + (n - 1)d = 255
⇒ 7 + (n - 1) × 8 = 255
⇒ 7 + 8n - 8 = 255
⇒ 8n - 1 = 255
⇒ 8n = 256
⇒ n = 32

Therefore, the required sum of the series = 322322[2 ∙ 7 + (32 - 1) ∙ 8]
= 16 [14 + 31 ∙ 8]
= 16 [14 + 248]
= 16 × 262 = 4192

Talk to Our counsellor