Question

the 5th and 10th term of ap are 18 and 38 respectively find ap

Answer 1

Answer:

let the 1st term of ap is a and common difference is d

so 5 th term =a+(5-1).d=18

=> a+4d=18

=> a=18-4d

again 10 th term =a+(10-1).d=38

=> a+9d=38

=> 18-4d+9d=38

=> 5d=38-18

=> d=20/5

=> d=4

so, a=18-4d=18-4.4=18-16=2

so the series is 2,2+4=6,6+4=10,10+4=14,14+4=18....

Answer 2

Answer:

Using our definitions:

a + 4d = 18

a + 9d = -2

subtract:

5d = -20

d = -4

mentally in a + 4d = 18, a = 34 <----- first term

so we want S(n) < 0, let's see when it is zero

(n/2)(2a + (n-1)d) = 0

n/2)(68 - 4(n-1)) = 0

n/2(68 - 4n + 4) = 0

n(36 - 2n) = 0

-2n^2 + 36n = 0

n^2 - 18n = 0

n = 0 or n = 18

Step-by-step explanation:

So when we add the first 18 terms we get zero, the terms are getting smaller

so when we add 19 terms we get a negative sum

PLEASE MARK ME AS A BRIANLIST