SUM OF FIRST N TERMS OF AN ARITHMETIC SERIES

Example 1 :

Find the sum of first 75 positive integers.

Solution :

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

The above series is an arithmetic series with last term 75.

Since we know the last term, the second formula can be used to find the required sum.

S_n = (n/2)(a₁ + l)

S₇₅ = (75/2)(1 + 75)

   = (75/2)(76)

= 75 x 38

= 2850

Example 2 :

Find the sum of first 125 natural numbers.

Solution :

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

The above series is an arithmetic series with last term 125.

Since we know the last term, the second formula can be used to find the required sum.

S_n = (n/2)(a₁ + l)

S₁₂₅ = (125/2)(1 + 125)

= (125/2)(1 + 125)

= (125/2)(126)

= 125 x 63

= 7875

Example 3 :

Find the sum of the terms in following arithmetic series.

6 + 5¼ + 4½ + ......... to 25 terms

Solution :

In the given arithmetic series a₁ = 6 and n = 25.

Common difference : 

d = a₂ - a₁

= 5¼ - 6

= 21/4 - 6

= (21 - 24)/4

= -3/4

Since we do not know the last term, the first formula can be used to find the required sum.

    Sn = (n/2)[2a₁ + (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

Example 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 :

Let n be the number of terms required in the given arithmetic sequence to get the sum 368.

S_n = 368

(n/2)[2a₁ + (n - 1)d] = 368 ----(1)

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

 a₁ = 60, 

d = a₂ - a₁

= 60 - 56

= -4

Substitute a₁ = 60 and d = -4 in (1).

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

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

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

Multiply both sides by 2.

  n[124 - 4n] = 736

124n - 4n² = 736

4n² - 124n + 736 = 0

Divide both sides by 4.

n² - 31n + 184 = 0

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

n = 23 or 8

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

Example 5 :

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

Solution :

The first 3 digit number is 100 and the last 3 digit number is 9999.

3 digit numbers which are divisible by 9 :

  108, 117, 126, ............ 999

3 digit numbers which are divisible by 9 form an arithmetic sequence.

a₁ = 108

d = a₂ - a₁

= 117 - 108

= 9

n = [(l - a₁)/d] + 1

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

= [891/9] + 1

= 99 + 1

= 100

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

= (100/2)[108 + 999]

= 50[1107]

= 55350

Example 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)[2a₁ + (11 - 1)d] = 44

2a₁ + 10d = (44 x 2)/11

2a₁ + 10d = 4 x 2

2a₁ + 10d = 8 ----(1)

Sum of next 11 terms = 55

Sum of first 11 terms + Sum of next 11 terms = 44 + 55

Sum of first 22 terms = 99

S₂₂ = 99

     (22/2)[2a₁ + (22 - 1)d] = 99

11[2a₁ + 21d = 99

2a₁ + 21d = 9

2a₁ + 21d = 9 ----(2)

(2) - (1) :

11d = 1

d = 1/11

Substitute d = 1/11 in (1).

2a₁ + 10(1/11) = 8

2a₁ + 10/11 = 8

2a₁ = 8 - 10/11

2a₁ = (88 - 10)/11

2a₁ = 78/11

a₁ = 39/11

a₂ = 39/11 + 1/11 = 40/11

a₃ = 40/11 + 1/11 = 41/11

Arithmetic series :

39/11 + 40/11 + 41/11 + ............

Example 7 :

Solve for x :

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

Solution :

The above series is an arithmetic series with

a₁ = 1, d = 5 and l = x

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

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

= (x - 1 + 5)/5

= (x + 4)/5

S_n = 148

(n/2)[a₁ + l] = 148

(1/2)(n)[a₁ + l] = 148

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

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

(x + 4)(x + 1) = 1480

x² + x + 4x + 4 = 1480

x² + 5x + 4 - 1480 = 0

x² + 5x - 1476 = 0

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

x = 36 or -41

The given arithmetic series is an increasing series with positive values. So, the last term can never be a negative value.

Therefore the last term of the series is 36.

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