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

Heya!

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Let,

First term = a
Common difference = d

Given,
Sum of the first three terms = 33

Second term = a + d
third term = a + 2d

33 = a + a + d + a + 2d
33 = 3a + 3d
33 = 3 ( a + d )
11 = a + d

d = 11 - a

Given,

Product of first and third term exceeds second term by 29

a1 × a3 = a2 + 29

a × a + 2d = a + d + 29

a² + 2 ad = a + d + 29

a² + 2ad - a - d = 29

a² + 2a ( 11 - a) - a - (11 - a) = 29

a² + 22a - 2a² - a - 11 + a = 29

- a² + 22a - 11 = 29

- a² + 22a - 11 - 29 = 0

- a² + 22a - 40 = 0

a² - 22a + 40 = 0

Splitting the middle term

a² - 20 a - 2a + 40 = 0

(a² - 20a) - (2a - 40) = 0

a ( a - 20) - 2 (a - 20) = 0

( a - 2) ( a -20) = 0

a - 2 = 0
a = 2

a - 20 = 0
a = 20


Case - 1

a = 2
d = 11 - a = 11 - 2 = 9

A.P = 2 , 11 , 20 ...

Case - 2

a = 20
d = 11 - a = 11-20 = -9

A.P = 20 , 11 , 2 ...


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Hope it helps...!!!

Answer 2

Hi

Here is your answer.