Given an arithmetic progression with Asub50=10 and Asub100 = -10. Find d and Ssub50

Question

Progression, Arithmetic progression.

Answer 1

Given artihmetic progression terms

a+49d = 10 ---> (1)

a+99d = -10 ---> (2)

To eliminate the a value subtract (1) from (2).

a+49d = 10

a+99d = -10

(-) (-)     (+)

___________

-50d = 20

Divide to each side by negitive 50.

d = -2/5

Substitute the d value in (1).

a+49(-2/5) = 10

a-98/5 = 10

a = 10+98/5

a = 148/5

n = 50

= 50/2[296/5+(49)(-2/5)]

= 25[296/5-98/5]

= 25[198/5]

= 990