If the sum of k term,3k²-k then find the 6th term of an AP.
Please help!!!:-)
Thanks.
Answers
Answered by
1
here is your answer
sum of k terms = k/2(2a + (k-1)d)
3k2 - k = k/2(2a + 6k - 6)
3k2 - k = k(a + 3k - 3)
By dividing both sides by k,
3k - 1 = a + 3k - 3
a = 3k - 3k -1 + 3
a = 2
So, the first term is 2.
Mark me as brainliest
sum of k terms = k/2(2a + (k-1)d)
3k2 - k = k/2(2a + 6k - 6)
3k2 - k = k(a + 3k - 3)
By dividing both sides by k,
3k - 1 = a + 3k - 3
a = 3k - 3k -1 + 3
a = 2
So, the first term is 2.
Mark me as brainliest
Similar questions