The fourth term of an arithmetic progression is one less than twice the
second term - If the sixth term is 7
find the first term

Let the first term of the AP be a and the common difference of the AP be d

We know the formula for finding the nth term,

Here,

By using formula,

According to given condition,

⇛ a4 = 2a2 - 1

⇛ a + 3d = 2(a + d) - 1

⇛ a + 3d = 2a + 2d - 1

⇛ a - 2a + 3d - 2d = -1

⇛ -a + d = -1

⇛ a - d = 1

Then,

⇛ a = d + 1-----------(1)

It is given that, a6 = 7

⇛ a + 5d = 7

Putting a from eq.(1),

⇛ d + 1 + 5d = 7

⇛ 6d + 1 = 7

⇛ 6d = 6

⇛ d = 6/6 = 1

Putting value of d in eq.(1),

⇛ a = 1 + 1 = 2

⛈ First term of the AP(a) = 2

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whats the question?

step-by-step explanation:

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0