# Determine the quadratic ax^2+bx+c, if a,b,c are terms of arithmetic sequence. a=2t-3,b=5t+1,c=4t-7

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### 2 Answers

We are given that in the quadratic equation ax^2+bx+c, a,b,c are terms of arithmetic sequence such that a = 2t-3, b = 5t+1 and c = 4t-7

As consecutive terms of an AP have a common difference:

4t - 7 - 5t - 1 = 5t + 1 - 2t + 3

=> -t - 8 = 3t + 4

=> 4t = -12

=> t = -3

a = 2t - 3 = -6 - 3 = -9

b = 5t + 1 = -15 + 1 = -14

c = 4t - 7 = -12 - 7 = -19

The quadratic equation is -9x^2 - 14x - 19 = 0

=> 9x^2 + 14x + 19 = 0

**The required quadratic equation is 9x^2 + 14x + 19 = 0**

If 2t-3, 5t+1, 4t-7, are the consecutive terms of an arithmetical progression, then the middle term is the arithmetical mean of its neighbor terms.

5t+1=(2t-3+4t-7)/2

We'll combine like terms inside brackets:

5t+1=(6t-10)/2

5t+1=2(3t-5)/2

5t+1=3t-5

We'll move the terms in t to the left side and the numbers alone to the right side:

5t-3t=-1-5

2t=-6

t=-6/2

t=-3

Since the coefficients a,b,c are depending on t, we'll determine them:

a = 2t - 3

a = -6 - 3

a = -9

b = 5t + 1

b = -15+1

b = -14

c = 4t - 7

c = -12 - 7

c = -19

The quadratic equation is: -9x^2 - 14x - 19 = 0

We'll multiply by -1 and we'll get:

**9x^2 + 14x + 19 = 0**