Question

Find the sum of the following series
5+(-41)+9+(-39)+13+(-37)+---+81+(-3)

Answer 1

Answer:

Step-by-step explanation:

The given series consists of two a.p series

5+9+13+...+81

and

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

In the first a.p

a=5 , d=4,an=81

∴5+(n-1)4=81

⇒(n-1)4=76

⇒n-1=19

⇒n=20

∴n=20

∴sn=20/2(5+81)

=10*86

=860

and in the second a.p

a=(-41)

d=2

an=(-3)

∴(-41) + (n-1)2=(-3)

⇒(n-1)2=38

⇒n-1=19

⇒n=20

∴sn=20/2[(-41)+(-3)]

=10*(-44)

=-440

∴total sum of the series

=860-440

=420

Answer 2

Answer:

Step-by-step explanation:

Solution :-

The series can be written as

(5 + 9 + 13 + .... + 81) + [(- 41) + (- 39) + (- 37) + (- 35) ... (- 5) + (- 3)]

For the series (5 + 9 + 13 + ... + 81)

a = 8

d = 9 - 5 = 5

and a(n) = 81

Then, a(n) = 5 + (n - 1)4 = 81

or, (n - 1)4 = 76

⇒ n = 20

S(n) = 20/2(5 + 81)

= 860

For Series (- 41) + (- 39) + (- 37) + (- 35) ... (- 5) + (- 3)

a(n) = - 3

a = - 41

d = 2

Then, a(n) = - 41 + (n - 1)(2)

⇒ n = 20

Now, S(n) = 20/2[- 41 + (- 3)]

= - 440

Sum of the series = 860 - 440

= 420

Hence, the sum of the following series is 420.