Sum of 1
st and 31st terms of an arithmetic sequence is 80.
Answers
Answer:
1st term is 10, 31st term is 70. Common difference is 2
Step-by-step explanation:
1st term = x
common difference = y
31st term = x + 30y
2x + 30y = 80
x + 15y = 40 (dividing both sides by 2)
x = 10
y = 2
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Answer:
As the value of n increases,the term increases and sum of n terms increases (n is always integer)
Sum of n terms
Min value=80,n=1
Max value=infinite,n=infinite
As n approaches infinite,sum of n term approaches infinite
Step-by-step explanation:
I’m assuming that the common difference, d, is -3, as there is no finite solution to the problem as posed.
With first term, a, =80, and d=-3, the 27th term is 80 + 26 x (-3)=2, and the 28th term is 80 + 27 x (-3) = -1.
Thus the sums increase up to n=27, and then decrease ever after.
The greatest sum is for n = 27, and the sum is
(n/2)*(2a + (n-1)d) = (27/2)*(2*80+26*(-3)) = 1107