Question

solve this problem
which is written in copy​

Answer 1

Answer:

4

Step-by-step explanation:

Given, AP is 25,22,19...116

First term, a = 25.

Common difference, d = -3.

Let the number of terms be n.

Given, Sn = 116.

Now,

Sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]

⇒ 116 = (n/2)[50 + (n - 1) * (-3)]

⇒ 232 = (n)[50 - 3n + 3]

⇒ 232 = n[53 - 3n]

⇒ 232 = 53n - 3n²

⇒ 53n - 3n² - 232 = 0

⇒ 3n² - 53n + 232 = 0

⇒ 3n² - 29n - 24n + 232 = 0

⇒ n(3n - 29) - 8(3n - 29) = 0

⇒ (n - 8)(3n - 29) = 0

⇒ n = 8, 29/3{∵not possible}

⇒ n = 8

∴ Last term t₈ = a + (n - 1) * d

                       = 25 + (8 - 1) * (-3)

                       = 25 - 21

                       = 4.

Therefore, Last term is 4.

Hope it helps!