Question

The sum of the first three terms of an AP is 33 .If the product of the first and the third term exceeds the second term by 29,find the AP

Answer 1

Answer:

the AP is

2, 11, 20 ........

Step-by-step explanation:

let the 3 terms be a-d, a, a + d

sum of the numbers is 33

a - d + a + d = 33

3a = 33

a = 33/3 = 11

product of first and last term = 2nd term + 29

(a - d)(a + d) = a + 29

${a}^{2} - {d}^{2} = a + 29 \\ {11}^{2} - {d}^{2} = 11 + 29 \\ 121 - {d}^{2} = 40 \\ - {d}^{2} = 40 - 121 \\ { - d}^{2} = - 81 \\ {d}^{2} = 81 \\ d = \sqrt{81} \\ d = 9$

therefore the numbers are

a - d = 11 - 9 = 2

a = 11

a + d = 11 + 9 = 20

the AP is

2, 11, 20

hope you get your answer

Answer 2

Answer:

heya✨✨

Step-by-step explanation:

Let the first three terms of AP is a-d, a, a+da−d,a,a+d.

According to problem,

a-d+a+a+d=33a−d+a+a+d=33 \rightarrow (1)→(1)

(a-d)(a+d)=a+29\rightarrow (2)(a−d)(a+d)=a+29→(2)

From (1)(1), we get

3a=333a=33

a=11a=11

From (2)(2), we get

a^2-d^2=a+29\rightarrow (3)a

2

−d

2

=a+29→(3)

Put a=11a=11 in equation (3)(3)

121-d^2=11+29121−d

2

=11+29

121-d^2=40121−d

2

=40

121-40=d^2121−40=d

2

81=d^281=d

2

d=\pm 9d=±9

When a=11a=11 & d=9d=9

A.P: 2,11, 20, 29,.....2,11,20,29,.....

When a=11a=11 & d=-9d=−9

A.P: 20, 11, 2,.....20,11,2,......