Question

If k+3, 2k+6 and 8 are three consecutive terms of an ap. find the value of K ?

Answer 1

if a,b,c are in AP then
$2b = a + c \\ 2(2k+ 6) = k + 3 + 8 \\ 4k + 12 = k + 11 \\ 3k = -1 \\ k = -\frac{1}{3}$
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Answer 2

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ansWer ⤵️⤵️⤵️

as these terms are in AP
so d = T2 - T1 = T3 - T2
2k + 6 - ( k+3) = 8 - (2k+6)
2k+6-k-3 = 8-2k-6
k+3 = -2k + 2
3k = -1
k = -1/3
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