Question

please solve it fast
class 10th​

Answer 1

Answer:

Let the 3 terms of an AP be a-d,a,a+d

a + a + d + a - d = 33

3a = 33

So, a = 11

Therefore, the terms are 11-d,11,11+d

(11-d)(11+d) = 29 + 11

121 + 11d -11d - d^2 = 40

121 - d^2 = 40

d^2 = 81

d = 9

Therefore the AP is 2, 11, 20

HOPE THAT THIS WAS HELPFUL !!!!

Step-by-step explanation:

Answer 2

Answer:

let the first term be A

the common difference be D

then the sum of first 3 term

a+a+ d+a+2d=3a+3d

given that 3(a+d)= 33

a+d=11

d=11-a

second term of AB be 11

the product of the first and third term = (a)(a+2d)=a(a+2(11-a)

=a(a+22-2a)

=a(22-a)

Given 22a-a2-29=a+d

22a-a2-29=11

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