Question

The sum of three consecutive terms of AP is 6 and thier product is -90.find their terms

Answer 1

EXPLANATION.

Sum of three consecutive terms of an A.P. = 6.

Products of three consecutive terms of an A.P. = -90.

As we know that,

Three consecutive terms,

⇒ a - d, a , a + d.

Sum of three consecutive term = 6.

⇒ a - d + a + a + d = 6.

⇒ 3a = 6.

⇒ a = 2.

Products of three consecutive term = -90.

⇒ (a - d)(a)(a + d) = -90.

⇒ (a - d)(a + d)(a) = -90.

As we know that,

Formula of : (a - b)(a + b) = (a² - b²).

⇒ (a² - d²)(a) = -90.

Put the value of a in this equation, we get.

⇒ [(2)² - d²](2) = -90.

⇒ [4 - d²](2) = -90.

⇒ 8 - 2d² = -90.

⇒ - 2d² = - 90 - 8.

⇒ - 2d² = - 98.

⇒ 2d² = 98.

⇒ d² = 98/2.

⇒ d² = 49.

⇒ d = √49.

⇒ d = ± 7.

As we know that,

Three consecutive term :

a - d, a, a + d.

First term = a = 2.

Common difference = d = 7.

Put the values in this equation, we get.

⇒ (2 - 7), 2, (2 + 7).

⇒ -5, 2, 9.

First term = a = 2.

Common difference = d = -7.

Put the values in this equation, we get.

⇒ (2 - (-7)), 2, (2 + (-7)).

⇒ 9, 2, -5.

                                                                                                                     

MORE INFORMATION.

Supposition of terms in A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.

(3) = Four terms as : a - 3d, a - d, a + d, a + 3d.

General terms of an A.P.

General term (nth term) of an A.P. is given by,

Tₙ = a + (n - 1)d.

Sum of n terms of an A.P.

Sₙ = n/2 [ 2a + (n - 1)d]  Or  Sₙ = n/2 [ a + Tₙ].

(1) = If sum of n terms Sₙ is given then general term Tₙ = Sₙ - Sₙ₋₁ where Sₙ₋₁ is sum of (n - 1) terms of an A.P.

Answer 2

Let the 3 consecutive numbers be a-d, a, and a+d respectively.

where a = first term of the AP and d=is their common difference.

Now given that sun of the three numbers is 6.

So we get =>

a + a-d + a + d = 6

=> 3a = 6

=>a = 2

Also, a(a-d)(a+d) = -9

Putting a = 2 we get =>

2(2-d)(2+d) = -9

By using (a+b)(a-b) = a²-b² we get =>

2(2² - d²) = -9

=> 2(4- d²) = -9

=> 8 -2d² = -17

=>-2d² = -25

=>d² = 25/2

=> d =√(25/2)

Hence the terms are =>

2 - √(25/2),2 and 2 + √(25/2) respectively.

Also we can write the terms as

2 - (5/√2) , 2 and 2 + (5/√2)

Remember

1)An = a +(n-1)d

where An = nth term of AP.

2)$n =\dfrac{n}{2}$

$[2a + (n-1)d].$

where Sn is the sum of n terms of AP.