Question

if k+1,3k and 4k+2 be any three consecutive terms of an Ap find the value of k

Answer 1

Heya !!!





AP = K + 1 , 3K and 4K+2





Here,




First term (T1) = K +1



Second term ( T2) = 3K





And,



Third term ( T3) = 4K +2





Common difference (D) = T2 - T1



=> 3K - ( K +1)



=> 3K - K -1



=> 2K -1




Also,



Common Difference = T3 - T2






=> 4K + 2 - ( 3K)




=> 4K + 2 - 3K




=> K +2




As we know that,



Common Difference of an AP is always equal.




So,




T2 - T1 = T3 - T2



2K - 1 = K +2




2K - K = 2 +1




K = 3.





★ ★ ★ HOPE IT WILL HELP YOU ★ ★ ★

Answer 2

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հίί ʍαtε_______✯◡✯
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һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________
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✯♦first term t1 = (k+1)

♦second term t2 = 3k

♦third term t3= (4k +2)
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given that t1, t2, t3 are in AP

so,

t2 - t1 = t3 - t2

3k - (k + 1) = (4k + 2) - 3k.

3k- k - 1 = 4k + 2k -3k.

2k - 1 = k + 2

2k = k +2+1

2k = k + 3

2k - k = 3

k = 3 .

$hence \: the \: value \: of \: k \: is \: 3 \\ \\ hope \: \: it \: helps$
!!!✌✌
#BRAINLY