Question

in an AP sum of first three consecutive terms is 15 and their product is 105 find the numbers​

Answer 1

Answer:

Let 3 consecutive digits be a-d,a,a+d.

sum of digits=15 (given)

a-d+a+a+d=15

3a=15

a=5.

product of digits =105. (given)

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

a^3-ad^2=105

put a=5

125-5d^2=105

5d^2=20

d^2=4

d=2.

hence our three numbers are 3,5,7

PLEASE MARK IT AS BRAINLIEST ANSWER.

Answer 2

let three number in AP be a - d , a and a + d

according to the question,,

the sum of three consecutive terms equals to 15

$⟹a \: - d \: + a + a + a + d \: = 15 \\ \\⟹ 3a \: = 15 \\ \\ ⟹a \: = 5$

and the product of three consecutive term is 105

$( ⟹\: a \: - \: d) \: (a) \: ( \: a \: + \: d) \: = 105 \\ \\⟹( 5 - d \: ) \: (5)(5 + \: d) = 105$

$⟹(25 - {d}^{2} )5 = 105 \\ \\⟹ 25 - {d}^{2}$

$= \frac{105}{5}$

$25 - d {}^{2} = 21 \ \tt \implies \: {d = 4}^{2}$

d ≠ 2

So, d = 2 then term in an AP are 5 - 2 , 5 and 5 +2 i.e 3,5,7

When d = -2 then the term in ap are 5 + 2 , 5 , 5-2 ie 7,5,3