Question

Find three numbers in AP whose sum is 15 and product of their extremes is 21​

Answer 1

Answer:

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Step-by-step explanation:

Let the required numbers be (a−d),a,(a+d).

Then,

a−d+a+a+d=21

3a=21

a=7

Also,

(a−d)a(a+d)=231

a(a

2

−d

2

)=231

7(49−d

2

)=231

7d

2

=112

d

2

=16

d=±4

Hence, the required numbers are (3,7,11) or (11,7,3).