Question

The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is *​

Answer 1

Answer:

from end, 1st term = –1000 & common difference = 5

so 10th term =

–1000+(10–1)×5 = —1000+45 = –955.

Ans.

Answer 2

Step-by-step explanation:

A= -5 , d=a-t2= -5)-(-10)=-5

N= 10

$t_{n} = a + (n - 1)d \\ = ( - 5) + (10 - 1) - 5 \\ = ( - 5) + (9 \times ( - 5) = \\ = ( - 5) + ( - 45) \\ t_{10} = - 40$