Question

Find three numbers in A.P. such that their sum is 36 and the third number exceeds the sum of first and second by 2.​

Answer 1

Answer:

7, 12,17

Step-by-step explanation:

Let the three terms of A.P. be a-d , a , a+d.

a-d+a+a+d = 36

3a = 36

a = 12

here first term - a-d

second term - a

third term - a+d

so according to the statement third term exceeds the sum of first and second term by 2

sum of first and second term = (a-d)+(a). third number exceeds by 2 = (a+d) + 2

(a-d)+(a) = (a+d) + 2

a-d+a = a+d+2

2a - d = a + d + 2

2a -a -2 = d +d

a - 2 = 2d. [ a=12]

12 - 2 = 2d

10 = 2d

d = 5

so first term = a-d

= 12-5

= 7

second term = a

= 12

third term = a+d

= 12 +5

= 17

So, the three numbers in A.P are 7,12,17

Answer 2

Answer = Thus, the numbers are 15,12 and 9.

Let the three numbers be a−d,a,a+d.

It is given that the sum of the numbers is 36, therefore,

a−d+a+a+d=36

⇒3a=36

⇒a=

3

36

=12......(1)

It is also given that the product of the numbers is 1620, therefore using equation 1 we have,

(a−d)(a)(a+d)=1620

⇒(12−d)(12)(12+d)=1620

⇒(12−d)(12+d)=

12

1620

⇒12

2

−d

2

=135

⇒d

2

=144−135

⇒d

2

=9

⇒d=−3,d=3

With a=12 and d=3, the three numbers are as follows:

a−d=12−3=9

a=12

a+d=12+3=15

Thus, the numbers are 9,12 and 15.

With a=12 and d=−3, the three numbers are:

a−d=12−(−3)=12+3=15

a=12

a+d=12−3=9

Thus, the numbers are 15,12 and 9.