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Pavel [41]
3 years ago
9

For what value of k will k+9,2k-1 and 2k+7 are the consecutive terms of an A.P

Mathematics
1 answer:
Stolb23 [73]3 years ago
8 0

k = 18

the common difference d of an arithmetic sequence is

d = a₂ - a₁ = a₃ - a₂ = ....

a₂ - a₁ = 2k - 1 - k - 9 = k - 10 and

a₃ - a₂ = 2k + 7 - (2k - 1) = 2k + 7 - 2k + 1 = 8, hence

k - 10 = 8 ⇒ k = 8 + 10 = 18

the 3 consecutive terms are

27, 35, 43


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\bf \qquad \qquad \textit{direct proportional variation}
\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------

\bf \textit{f(x) varies directly with }x^2\qquad f(x)=kx^2
\\\\\\
\textit{we also know that }
\begin{cases}
f(x)=96\\
x=4
\end{cases}\implies 96=k(4)^2\implies 96=16k
\\\\\\
\cfrac{96}{16}=k\implies 6=k\qquad therefore\qquad \boxed{f(x)=6x^2}
\\\\\\
\textit{now, when }\stackrel{f(2)}{x=2}\textit{ what is \underline{f(x)}?}\qquad f(2)=6(2)^2
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