If 2k,k+10 and 3k+2 are three consecutive terms of an A.P.,then find the value of k
Answers
Answered by
6
Answer:
This do not forms an AP as d(common difference) is not same.
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Answered by
26
Answer:
k=6.
Explanation:
From the question we get that the a1= 2k, a2= k+10, a3= 3k+2 are in arithmetic progression so we know that if three consecutive terms are in A.P then we can write that b-a=c-b which is second term - first term = third term - second term.
Hence, on substituing the values from the question we will get that K+10 - 2k = 3k+2-(k+10).
K+10-2k =3k+2-k-10.
k -2k +10 = 3k-k +2-10.
-k +10= 2k -8.
2k+k= 10+8.
Which on solving we will get that the value of k will be.
k =18/3= 6.
Hence, k= 6.
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