**Sum of arithmetic progression formula :**

An arithmetic series is a series whose terms form an arithmetic sequence.

We use the one of the formula given below to find the sum of arithmetic series.

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

**Sn = (n/2) [a + l]**

- "n" stands for the total number terms
- "a" stands for the first term
- "d" stands for common difference
- "l" stands for the last term

We can use the first formula to find sum of n terms, if we don't have the last term of the given arithmetic series.

We can use the second formula to find sum of n terms, if we have the last term of the given arithmetic series.

**Problem 1 :**

Find the sum of first 75 positive integers

**Solution :**

1 + 2 + 3 + ..........+ 75

Since we know the last term, we can use the second formula to find the sum of the arithmetic series.

Sn = (n/2) (a+l)

= (75/2) (1+75)

= (75/2) (76)

= 75 x 38

= 2850

**Problem 2 :**

(ii) 125 natural numbers

**Solution :**

1 + 2 + 3 + ..........+ 125

Total number of terms in the series is 125 so n = 125

Sn = (n/2) (a+L)

= (125/2) (1+125)

= (125/2) (126)

= 125 x 63

= 7875

**Problem 3 :**

6 + 5 ¼ + 4 ½ + ......... 25 terms

**Solution :**

In this series we have 25 terms so n = 25

a = 6 d = t₂ - t₁ ==> = 5 ¼ - 6 ==> -3/4

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

S₂₅ = (25/2) [ 2(6) + (25-1) (-3/4) ]

= (25/2) [ 12 + (24) (-3/4) ]

= (25/2) [ 12 + (6) (-3) ]

= (25/2) [ 12 - 18 ]

= (25/2) [ -6 ]

= (25) [ -3 ]

= -75

**Problem 4 :**

In an arithmetic sequence 60,56,52,48,....... starting from the first term,how many terms are needed so that their sum is 368?

**Solution :**

In this problem we have to find the number of terms are needed so that their sum is 368

S n = 368

60,56,52,48,.......

a = 60 d = 56 - 60 = -6

(n/2) [ 2a + (n-1) d ] = 368

(n/2) [ 2(60) + (n-1)(-4) ] = 368

(n/2) [ 120 - 4 n + 4 ] = 368

(n/2) [ 124 - 4 n ] = 368

n [ 124 - 4 n ] = 368 x 2

n [ 124 - 4 n ] = 368 x 2

124 n - 4 n² = 736

4 n² - 124 n + 736 = 0

÷ by 4 = > n² - 31 n + 184 = 0

(n-23) (n-8) = 0

n = 23,8

Sum of 8 or 23 terms of arithmetic sequence is 368.

**Problem 5 :**

Find the sum of all 3 digit natural numbers,which are divisible by 9.

**Solution :**

3 digit number starts from 100 and ends with 999. From this sequence we have to find number of terms which are divisible by 9 and also we have to find their sum.

108, 117, 126,........................999

The first number which is divisible by 9 from this sequence is 108, the second number will be 117 and the last number will be 999.

108 + 117+ 126 + ........... + 999

a = 108 d = 117 - 108 = 9 and L = 999

to find number of terms, we have to use the formula for (n)

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

n = [(999-108)/9] + 1

n = [891/9] + 1

n = 99 + 1

n = 100

S n = (n/2) [a+l]

= (100/2) [108 + 999]

= 50 [1107]

= 55350

Let us see the next example on "Sum of arithmetic progression formula"

**Problem 6 :**

In an arithmetic series,the sum of first 11 terms is 44 and the that of the next 11 terms is 55. Find the arithmetic series.

**Solution :**

Sum of first 11 terms = 44

S₁₁ = 44

(11/2)[2a + (11-1) d] = 44

2a + 10 d = (44 x 2)/11

2a + 10 d = 4 x 2

2 a + 10 d = 8 ----- (1)

Sum of next 11 term = 55

S₂₂ = S₁₁ + 55

S₂₂ = 44 + 55

S₂₂ = 99

(22/2)[2a + (22-1) d] = 99

2a + 21 d = (99 x 2)/22

2a + 21 d = 9

2 a + 21 d = 9 ----- (2)

Solving (1) and (2) we get

-11 d = -1

d = 1/11

Substitute d = 1/11 in the first equation we get

2 a + 10 (1/11) = 8

2 a + 10/11 = 8

2 a = 8 - (10/11) ==> 78/11 ==> a = 39/11

Therefore the series is (39/11) + (40/11) + (41/11) + ............

Let us see the next example on "Sum of arithmetic progression formula"

**Problem 7 :**

Solve 1 + 6 + 11 + 16 + .......... + x = 148

**Solution :**

148 represents sum of x number of terms starting from 1.

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

a = 1 d = 5

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

= [(x - 1)/5] + 1 ==> (x - 1 + 5)/5 ==> (x + 4)/5

148 = [(x + 4)/5] /2 [1+x]

148 = (x+4)/10 [1+x]

148 x 10 = (x + 4) (1 + x)

1480 = x + x² + 4 + 4x

1480 = x² + 4 + 5 x

x² + 5 x + 4 - 1480 = 0

x² + 5 x - 1476 = 0

(x - 36) (x + 41) = 0

x = 36,-41

Therefore the last term of the series is 36.

After having gone through the stuff given above, we hope that the students would have understood "Sum of arithmetic progression formula".

Apart from the stuff given above, if you want to know more about "Sum of arithmetic progression formula", please click here

Apart from the stuff "Sum of arithmetic progression formula" given in this section, if you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

**WORD PROBLEMS**

**HCF and LCM word problems**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**