Mathematics
Determine the value of k for which k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are in A.P.
AP
39 Likes
Answer
Since, k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are in A.P.
Hence, difference between consecutive terms are equal.
∴ 2k2 + 3k + 6 - (k2 + 4k + 8) = 3k2 + 4k + 4 - (2k2 + 3k + 6)
⇒ 2k2 - k2 + 3k - 4k + 6 - 8 = 3k2 - 2k2 + 4k - 3k + 4 - 6
⇒ k2 - k - 2 = k2 + k - 2
⇒ k2 - k2 + k + k = -2 + 2
⇒ 2k = 0
⇒ k = 0.
Hence, k = 0.
Answered By
23 Likes
Related Questions
Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
If numbers n - 2, 4n - 1 and 5n + 2 are in A.P., find the value of n and its next two terms.
State, true or false: if a, b and c are in A.P. then :
(i) 4a, 4b and 4c are in A.P.
(ii) a + 4, b + 4 and c + 4 are in A.P.
An A.P. consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.