Question

The sum of three terms in an AP is 9root2 If the sum of their squares is 118, then the product of these
three terms is

Answer 1

Answer:

Product of these three terms = -42root2

Step-by-step explanation:

Let the three terms be a-d, a, a+d.

So, sum of three terms = (a-d) + (a) + (a+d)

= a - d + a + a + d

= 3a

Since, the sum of three terms is 9root2.

So, 3a = 9root2

a = 3root2

Sum of their squares = (a-d)2 + (a)2 + (a+d)2

= (a2 + d2 - 2ad) + (a2) + (a2+d2+2ad) = 3a2 + 2d2

Since, the sum of their squares is 118.

So, 3a2 +2d2 = 118

3(3root2)2 + 2d2 = 118

3*9*2 + 2d2 = 118

54 + 2d2 = 118

2d2 = 118 - 54

2d2 = 64

d2 = 32

d = root32

d = 4root2

So, product of these terms = (a-d) (a) (a+d)

= (a2 - d2)(a)

= [(3root2)2 - (4root2)2](3root2) = [(9*2) - (16*2)] (3root2)

= [(18) - (32)] (3root2)

= (- 14)(3root2)

= (-42root2).

Answer 2

SOLUTION

GIVEN

• The sum of three terms in an AP is 9√2

• The sum of their squares is 118

TO DETERMINE

The product of these three terms

EVALUATION

Since three given terms are in AP

Let the terms are

a - d , a , a + d

Here it is given that the sum of these terms is 9√2

$\sf{(a - d) + a + (a + d) = 9 \sqrt{2} }$

$\implies \sf{3a = 9 \sqrt{2} }$

$\implies \sf{a = 3 \sqrt{2} }$

Again it is also stated that the sum of their squares is 118

$\therefore \: \sf{ {(a - d)}^{2} + {a}^{2} + {(a + d)}^{2} = 118 }$

$\sf{ \implies {(a - d)}^{2} + {(a + d)}^{2}+ {a}^{2} = 118 }$

$\sf{ \implies {a}^{2} - 2ad + {d}^{2} +{a}^{2} + 2ad + {d}^{2} + {a}^{2} = 118 }$

$\sf{ \implies \: 3 {a}^{2} + 2 {d}^{2} = 118 }$

$\sf{ \implies \: 3 {(3 \sqrt{2}) }^{2} + 2 {d}^{2} = 118 }$

$\sf{ \implies \:54 + 2 {d}^{2} = 118 }$

$\sf{ \implies \:2 {d}^{2} = 64 }$

$\sf{ \implies \: {d}^{2} = 32 }$

$\sf{ \implies \:d = 4 \sqrt{2} }$

Hence the numbers are

$\sf{ - \sqrt{2} \: , \: 3 \sqrt{2} \: ,7 \sqrt{2} }$

Therefore the product of these numbers

$= \sf{ - \sqrt{2} \: \times \: 3 \sqrt{2} \: \times \: 7 \sqrt{2} }$

$= \sf{ - 42 \sqrt{2} }$

FINAL ANSWER

The product of these three terms

$= - \sf{42 \sqrt{2} }$

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