Question

If the roots of the cubic x3 - 6x2 - 24x + c = O are the first 3 terms of an A.P., then sum to first 10 terms of the increasing A.P. is

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The sum of the ten terms is 290 and -250.

Step-by-step explanation:

Let ( a - d ), a, and (a +d) are the roots of the cubic polynomial.

$x^{3} -6x^{2} -24x + c = 0$

Sum of roots = -b/a  = -(-6)/1 = 6

Now, a - d + a + a + d = 6

3a = 6

a = 6/3 = 2

Product of roots = c/a = -24

(a -d)a + (a + d)a + (a -d)(a +d) = -24

$a^{2} - ad +a^{2} -ad +a^{2} -d^{2} = -24$

$3a^{2} -d^{2} = -24$

Putting the value of a = 2, we have

$3(2)^{2} - d^{2} = -24$

$d^{2} = 36^{2} \\d = 6, -6$

The Sum of n terms = $\frac{n}{2} [2a + (n-1)d]$

Putting the value of a and d = 6, then

The Sum of n terms =  $\frac{n}{2} [2(2) + (n-1)(6)] = n [3n -1]$

Putting the value of a and d = -6, then

The Sum of n terms = $\frac{n}{2} [2(2) + (n-1)(-6)] = n [5 - 3n]$

So, Sum of the first 10 terms = n (3n -1) = 10(3×10 - 1) = 10 (30 - 1) = 10 (29) = 290.

OR

Sum of the first 10 terms = n( 5 - 3n) = 10 ( 5 - 3×10) = 10 ( 5 - 30 ) = 10 (-25) = -250.

Therefore, the sum of the ten terms is 290 and -250.

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