Question

Find three consecutive terms of an ap whose sum is 21 and the product of the last two terms 63

Answer 1

Answer:

5, 7 and 9

Step-by-step explanation:

let the numbers in ap be a-d, a and a+d.

so,

a-d+a+a+d=21. (add all 3 terms as shown.

3a=21

a=7

now,

as given,

a(a+d)=63

7(7+d)=63

49+7d=63

7d=14

d=2

so after putting the values of a and d, we get the answer 5, 7 and 9 or 9, 7 and 5.

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Answer 2

let the consecutive numbers be (x), (x+1),(x+2)

its sum,

(x) +(x+1)+(x+2)=21

3x+3=21

3(x+1)=21

x+1=21/3

x=6

By putting value of x in each

we get

x=6

x+1=7

x+2=8

no. are 6,7,8

hope it will help you