The sum of 16 terms of an AP is -504,while the sum of its 9 terms is -126.find the sum of its 30 terms

Answer 1

The sum of n terms of an AP:

Sn = ( n / 2 ) ∙ [ 2 a1 + ( n - 1 ) ∙ d ]

The sum of 16 terms of an AP:

S16 = ( 16 / 2 ) ∙ [ 2 a1 + ( 16 - 1 ) ∙ d ]

S16 = 8 ∙ [ 2 a1 + 15 d ] = - 504

8 ∙ 2 a1 + 8 ∙ 15 d = - 504

16 a1 + 120 d = - 504

The sum of 9 terms of an AP:

S9 = ( 9 / 2 ) ∙ [ 2 a1 + ( 9 - 1 ) ∙ d ]

S9 = 4.5 ∙ [ 2 a1 + 8 d ] = - 126

4.5 ∙ 2 a1 + 4.5 ∙ 8 d = - 126

9 a1 + 36 d = - 126

Now you must solve system:

16 a1 + 120 d = - 504

9 a1 + 36 d = - 126

Try it.

The solutions are: a1 = 6 , d = - 5

The sum of 30 terms of an AP :

S30 = ( 30 / 2 ) ∙ [ 2 a1 + ( 30 - 1 ) ∙ d ]

S30 = 15 ∙ [ 2 ∙ 6 + 29 ∙ ( - 5 ) ]

S30 = 15 ∙ ( 12 - 145 ) = 15 ∙ ( - 133 ) = - 1995

Answer 2

I want know how u got the ,6 and-5

Answer 3

Workin for common difference and first term

Answer 4

Please where's the working for how you got the first term and the common difference??

Answer 5

The 9TH AND THE 22ND term of ap and 29 and 25 respectively. Find the sum of its first 60term