Sum of Arithmetic Series

Sum of Arithmetic Series

 Let us consider an arithmetic series,

a + (a + d) + (a + 2d) + (a + 3d) + ……. + (l – 2d) + (l – d) + l

Here, first term = a, common difference = d, last term (t_n) = l and the term before last term = l – d

If the sum of n terms is denoted by S_n, then

S_n = a + (a + d) + (a + 2d) + ……. + (l – 2d) + (l – d) + l ------------------------ (i)

Writing the terms in reverse order,

S_n = l + (l – d) + (l – 2d) + ……. + (a + 2d) + (a + d) + a ------------------------- (ii)

Adding the corresponding terms of (i) and (ii),

S_n                  = a + (a + d) + (a + 2d) + ……. + (l – 2d) + (l – d) + l

S_n         = l + (l – d) + (l – 2d) + ……. + (a + 2d) + (a + d) + a

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2S_n       = (a + l) + (a + l) + (a + l) + ……… + (a + l) + (a + l) + (a + l)

2S_n       = n times (a + l)

2S_n       = n (a + l)

S_n         = n/2 (a + l)

1. Find the sum of first 50 terms of the sequence 1, 3, 5, 7, 9, …..

Solution,

Here,    a = 1, d = 2 and n = 50

Now, we know,

S_n         = n/2 {2a + (n – 1) d}

S₅₀               = 50/2 {2×1 + (50 – 1) 2}

            = 25 {2 + 49 × 2}

            = 25 × 100

            = 2500  

2.  From the series 5 + 10 + 15 + …… + 45, find n then find S_n.

Solution,

Here,    a = 5, d = 5 and last term (t_n) = 45

We know,

            t_n          = a + (n – 1) d

            45        = 5 + (n – 1) 5

            (n – 1) = 8

Therefore, n      = 9

Now,    S_n         = n/2 (a + l)

                        = 9/2 (5 + 45)    = 9 × 25            = 225

3.  Find the first term of A.P. whose common difference is – 7 and the sum of 8 terms is – 76.

Solution, 

Sum of 8 terms (S₈) = − 76

Common difference (d) = − 7

First term (a) =?

We know,

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

            S₈         = 8/2 [2a + (8 – 1) × (− 7)]

−76      = 4 [2a + (-49)]

-19       = 2a – 49

30        = 2a

Therefore,        a = 15

4.  If the first term of A.P. is 100 and sum of its 10 term is 550, find the common difference.

Solution,

Here,    1^st term (a) = 100

Sum of its first 10 terms (S₁₀) = 550

We know that,

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

            S₁₀        = 10/2 [2 × 100 + (10 – 1) d]

            550      = 5 [200 + 9d]

            110      = 200 + 9d

            9d        = −90

Therefore, d = − 10

5.  Find the number of terms in an arithmetic series if its first term is 3, common difference is 4 and the sum of terms is 210.

Solution,

Here,    first term (a) = 3

Common difference (d)= 4

Sum of terms (S_n) = 210

i.e.       n/2 [2a + (n – 1) d]        = 210

or,        n/2 [ 2× 3 + (n – 1) 4]    = 210

or,        n/2 (6 + 4n – 4) = 210

or,        n/2 (4n + 2)                  = 210

or,        n/2 × 2 (2n + 1)             = 210

or,        n (2n + 1)                      = 210

or,        2n² + n – 210                = 0

or,        2n² + 21n – 20n – 210   = 0

or,        n (2n + 21) – 10 (2n + 21)         = 0

or,        (2n + 21) (n – 10)          = 0

therefore,         n = (- 21)/2 or 10

but n cannot be negative and fraction as it denotes number of terms.

So, n = 10

6.  Find the sum of first 12 terms of an A.P. whose 5^th term is 17 and the 10^th term is 42.

Solution,

Here,    5^th term = 17

i.e.        a + 4d   = 17

or,        a = 17 – 4d ------------------(i)

again,   10^th term = 42

i.e.       a + 9d = 42

or,        a = 42 – 9d ------------------(ii)

now, from (i) and (ii),

17 – 4d = 42 – 9d

or,        9d – 4d = 42 – 17

or,        d = 25/5 = 5

putting the value of d in (i),

            a = 17 – 4 × 5 = 17 – 20 = − 3

Now, sum of first 12 terms        = n/2 [2a + (n – 1) d]

                        = 12/2 [2 × (-3) + (12 – 1) 5]

                        = 6 [-6 + 55]     = 6 × 49          

                        = 294

7.  Find the sum of the numbers from 1 to 100 which are exactly divisible by 3.

Solution,

The sequence of the numbers from 1 to 100 which are exactly divisible by 3 is 3, 6, 9, ……. 99.

Here,    first term (a) = 3

            Common difference (d) = 6 – 3 = 3

Last term (t_n) = 99

i.e.       a + (n – 1) d      = 99

or,        3 + (n – 1) 3      = 99

or,        3n        = 99

or,        n = 99/3 = 33

Now, required sum (S_n) = n/2 (a + l)

                                    = 33/2 (3 + 99)

                                    = 33/2 × 102     = 1683

8.  The sum of the three numbers in A.P. is 15 and the sum of their squares is 83, find the numbers.

Solution,

Let the three numbers in A.P. be a, (a + d) and (a + 2d)

Now,    sum of the numbers = 15

i.e.        a + (a + d) + (a + 2d) = 15

or,        3a + 3d = 15

or,        a + d =

or,        a = 5 – d -----------(i)

Again, a² + (a + d)² + (a + 2d) ² = 83

or,         (5 – d)² + (5 – d + d)² + (5 – d + 2d)²     = 83

or,        25 – 10d + d² + 25 + 25 + 10d + d²         = 83

or,        75 + 2d²            = 83

or,        2d²       = 83 – 75

or,        d²         = 8/2 = 4

therefore, d       = (± 2)

Keeping the value of d in (i),

For d = 2

            a = 5 – d = 5 – 2 = 3

first number (a) = 3

second number (a + d) = 3 + 2 = 5

third number (a + 2d) = 3 + 4 = 7

for d = - 2

            a = 5 + 2 = 7

first number (a) = 7

second number (a + d) = 7 – 2 = 5

third number (a + 2d) = 7 – 4 = 3

Hence, three numbers in A.P. are 3, 5, 7 or 7, 5, 3.