Question

How many terms are there in AP whose first and fifth terms are -14 and 2 respectively and sum of the terms is 40

Answer 1

Answer:

10 terms

Step-by-step explanation:

T1 = -14

T5 = 2 = T1 + 4d => 2 = -14 + 4d => d= 4

Sn = n/2 [2T1 + (n-1)d]

40 = n/2 [2(-14) + (n-1)4]

80 = n(-28 + 4n - 4)

80 = -28n + 4n² -4n

4n² - 32n - 80 = 0

n² - 8n - 20 = 0

n² - 10n + 2n - 20 = 0

n(n-10) + 2(n-10) = 0

(n-10) (n+2) = 0

Therefore, n = 10 [Since, n (— N]

Hope it helps you..........