Question

the sum of the first three terms of an AP is 33.if the product of the first term and third term exceeds the 2nd terms by 29 .then find the AP​

Answer 1

Answer:

Let the terms in AP be (a - d), a, (a + d).

Sum of the three terms = a - d + a + a + d = 3a

3a = 33

a = 11

Also, from the given information, we have:

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

a2 - d2 = 11 + 29 = 40

121 - d2 = 40

d2 = 81

d = 9

Thus, the AP is

2, 11, 20, … or 20, 11, 2, …

Answer 2

Answer:

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

From First Condition:

a-d+a+a+d=33

3a=33

a=11-------(1)

From Second Condition

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

a^2+ad-ad-d^2=11+29

a^2-d^2=40

121-d^2=40

d^2=121-40

d^2=81

d=+-9

For a=11,d=9

A.P is 2,11,20

For a=11,d=-9

A.P is 20,11,2

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