Math, asked by krishnarajbiradar123, 1 year ago

For what value of K, 2K-7 , K+5 AND 3K+2 are three consecutive terms of an AP

Answers

Answered by sodhanimudit

Answer:

2/5

Step-by-step explanation:

Answered by atharvvtiwari

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

krishnarajbiradar123: The sum of first n terms of an ap is given by 3n²+5n find the common difference

atharvvtiwari: Since sum of first n terms = Sn = 3n ^ 2 + 5n,

atharvvtiwari: S1 = a1 = 3 ( 1 ) ^ 2 + 5 ( 1 ) = 3 + 5 = 8

atharvvtiwari: S2 = a1 + a2 = 3 ( 2 ) ^ 2 + 5 ( 2 ) = 12 + 10 = 22

atharvvtiwari: Therefore, ( a1 + a2 ) - a1 = 22 - 8

atharvvtiwari: Thus, a2 = 14

atharvvtiwari: Common difference = a2 - a1 = 14 - 8 = 6

atharvvtiwari: Done :-)

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