Question

Please help me solve these last 2 questions

Answer 1

let the AP be a-d, a, a+d

sum is 15

a-d+a+a+d=15

3a=15

a=5

product is 105

(a-d)a(a+d)=105

(a²-d²)a=105

a³-ad²=105____(1)

put a=5 in (1)

5³-5d²=105

125-5d²=105

-5d²=105-125

-5d²=-20

d²=-20/-5

d²=4

d=2

therefore the AP is 3,5,7

Answer 2

Answer:

Step-by-step explanation:

79.

Let no.s to be

a-d,a,a+d

a-d+a+a+d=15

3a=15

a=5

Also products of their is 105

Putting value of a

We get,

(5-d)(5)(5+d)=105

5^2-d^2=21

-d^2=21-25

-d^2=-4

d=2

So the required numbers r

5-2,2,5+2

3,2,7

80. Four terms r

a-3d+a-d+a+d+a+3d=20

4a=20

a=5

Product of extreme r 16

(a-3d)(a+3d)=16

5-3d*5+3d=16

25-9d^2=16

-9d^2=-9

-d^2=-1

d=1

Required no,are

2,4,6,8

Hope you will get

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