For what value of K, 2K-7 , K+5 AND 3K+2 are three consecutive terms of an AP
Answers
Answered by
0
Answer:
2/5
Step-by-step explanation:
Answered by
9
Answer:
k = 5
Step-by-step explanation:
Since 2k - 7, k + 5 and 3k + 2 are three consecutive terms of an AP,
( 3k + 2 ) - ( k + 5 ) = ( k + 5 ) - ( 2k - 7 )
The reason for doing this is that both the LHS and RHS are equal to the common difference of the AP.
Thus,
3k + 2 - k - 5 = k + 5 - 2k + 7
2k - 3 = - k + 12
3k = 15
k = 5
krishnarajbiradar123:
One more question
The sum of first n terms of an ap is given by 3n²+5n find the common difference
Since sum of first n terms = Sn = 3n ^ 2 + 5n,
S1 = a1 = 3 ( 1 ) ^ 2 + 5 ( 1 ) = 3 + 5 = 8
S2 = a1 + a2 = 3 ( 2 ) ^ 2 + 5 ( 2 ) = 12 + 10 = 22
Therefore, ( a1 + a2 ) - a1 = 22 - 8
Thus, a2 = 14
Common difference = a2 - a1 = 14 - 8 = 6
Done :-)
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