If a+1,2a+1,4a-1 are in A.P, then the value of a is

Mathematics

1 answer:

Digiron [165]3 years ago

6 0

Answer:

a = 2

Step-by-step explanation:

In an AP the difference between consecutive terms is common (equal), thus

t₂ - t₁ = t₃ - t₂ , that is

2a + 1 - (a + 1) = 4a - 1 - (2a + 1) ← distribute and simplify both sides

2a + 1 - a - 1 = 4a - 1 - 2a - 1

a = 2a - 2 ( subtract 2a from both sides )

- a = - 2 ( multiply both sides by - 1 )

a = 2

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F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$