Question

Find the sum of the arithmetic series:

$5 + ( - 41) + 9 + ( - 39) + 13 + \\ ( - 37) + 17 + ... + ( - 5) + 81 + \\ ( - 3).$

Class 10


Arithmetic Progressions

Answer 1

The Question has Two Series which are in A.P

1st Series : 5 + 9 + 13 + 17 + . . . . + 81

2nd Series : (-41) + (-39) + (-37) + . . . . . + (-3)

Let us Take the 1st Series :

5 + 9 + 13 + 17 + . . . . + 81

We can notice that the 1st term is 5 ⇒ a = 5

Common Difference = 2nd term - 1st term = 9 - 5 = 4

nth term = 81

We know that nth term = a + (n - 1)d = 81

⇒ 5 + (n - 1)4 = 81

⇒ 5 + 4n - 4 = 81

⇒ 4n = 80

⇒ n = 20

Sum of 'n' terms in an AP is given by : n/2 × { a + a + (n - 1)d }

but a + (n - 1)d = 81 and a = 5 and n = 2

⇒ Sum of the 20 terms of 1st series = 20/2 × {5 + 81} = 10 ×86 = 860

Now let us take the 2nd series :

(-41) + (-39) + (-37) + . . . . . + (-3)

We can notice that the 1st term is -41 ⇒ a = -41

Common Difference = 2nd term - 1st term = -39 - (-41) = 41 - 39 = 2

nth term = -3

We know that nth term = a + (n - 1)d = -3

⇒ -41 + (n - 1)(2) = -3

⇒ -41 + 2n  - 2 = -3

⇒ 2n = 43 - 3 = 40

⇒ n = 20

Sum of 'n' terms in an AP is given by : n/2 × { a + a + (n - 1)d }

but a + (n - 1)d = -3 and a = -41 and n = 20

Sum of 20 terms of 2nd series = 20/2 × {-41 + -3} = 10 × (-44) = -440

Sum of the Total Series = Sum of 20 terms of 1st Series + Sum of 20 terms of 2nd Series

⇒ Sum of the Given arithmetic Series = 860 - 440 = 420

Answer 2

Let,

S = 5 + (-41) + 9 + (-39) + 13 + (-37) + .........................(-5) + 81 + (-3)

This  series  contents two A.P. Series, making them in  sequence

S = 5 + 9 + 13 + ...........81 + (-41) + (-39) + (-37) ............ (-5) + (-3)

S = S₁ + S₂ (say)  ..............................i)

Where,

S₁ =  5 + 9 + 13 + ...........81

S₂ = (-41) + (-39) + (-37) ............ (-5) + (-3)

Now,

Find S₁:

S₁ = 5 + 9 + 13 + ...........81

First term (a) = 5

Last term (l) = 81

Let the  number of  terms  be  n  and  common difference d,

d = 9 - 5 = 13 - 9 = 4

We know that ,

l  = a + (n-1)d

81 = 5 + (n-1)4

76 = (n-1)4

n-1 = 19

n = 20

Again,

Sum of  terms = n/2[a+l]

                       = 20/2[5+81]

                       = 10[86]

                      S₁ = 860

Find S₂:

S₂ = (-41) + (-39) + (-37) ............ (-5) + (-3)

First term (a) = -41

Last term (l) = -3

Let the  number of  terms  be  n  and  common difference d,

d = -39 - (-41) = -37 - (-39) = 2

We know that ,

l  = a + (n-1)d

-3 = -41 + (n-1)2

38 = (n-1)2

n-1 = 19

n = 20

Again,

Sum of  terms = n/2[a+l]

                       = 20/2[-41+(-3)]

                       = 10[-44]

                      S₂ = -440

Hence,

S = S₁ + S₂

S = 860 + (-440)

S = 420