In an A.P the first term is I and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120? (1) 6 (2) 7 (3) 8 (4) 9
Answers
Answered by
0
Answer:
3) n = 8
Step-by-step explanation:
a = 1
d = 4
Sn = 120
Sn = n/2[2a +(n-1)d]
120 = n/2 [2×1 + (n-1)4]
120 = n/2 [2 + 4n - 4]
120 = n/2 [4n - 2]
120×2 = n(4n - 2)
240 = 4n² - 2n
4n² - 2n - 240 = 0
2(2n² - n - 120) = 0
2n² - n - 120 = 0
2n² - 16n + 15n - 120 = 0
2n (n - 8) + 15 (n - 8) = 0
(2n + 15) (n-8) = 0
n = -15/2 , 8
n can't be in fraction. Therefore, n is 8.
Answered by
2
Answer:
8
Step-by-step explanation:
The value of n is 8. So, 8 terms of the A.P must be taken for their sum to be equal to 120.
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