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

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...!!!