Question

Find sumof given

a.p. 5+(-41)+9...........81+(-3)

Answer 1

The answer is given below :

The AP is

5 + (-41) + 9 + ... + 81 + (-3)

We can write the AP as the sum of two different APs.

Then,

(5 + 9 + ... + 81) + {(-41) + (-39) + ... + (-3)}

We take the first AP.

Then,

5 + 9 + ... + 81

The first term of the AP = 5 and
common difference = 4.

The last term = 81

=> first term + (n - 1)×common difference, where n is the total number of terms in the AP

=> 5 + (n - 1)×4 = 81

=> 5 + 4n - 4 = 81

=> 4n + 1 = 81

=> 4n = 81 - 1

=> 4n = 80

=> n = 80/4

=> n = 16

So, the first AP has 16 terms.

Then, the sum of the first series is

= (n/2) × (first term + last term)

= (16/2) × (5 + 81)

= 8 × 86

= 688

The second AP is taken as

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

The first term of the AP is (-41) and
common difference = 2.

The last term = -3

=> first term + (m - 1)×common difference = -3, where m is the total number of terms in the AP.

=> (-41) + (m - 1)×2 = (-3)

=> - 41 + 2m - 2 = - 3

=> 2m - 43 = - 3

=> 2m = - 3 + 43

=> 2m = 40

=> m = 40/2

=> m = 20

So, the second AP has 20 terms.

Then, the sum of the 2nd AP

= (m/2) × (first term + last term)

= (20/2) × ( - 41 - 3)

= 10 × (-44)

= - 440

Therefore, the sum of the given AP

5 + (-41) + 9 + ... + 81 + (-3) is

= 688 - 440

= 248

Thank you for your question.