Question

the sum of the first three terms of an A.P is 33. If the product of first and the third and the third term exceeds the second term by29, find the A.P

Answer 1

Let the first term be a ,
The common difference be d .
Then the sum of first three terms = a+a+d+a+2d = 3a+3d .

Given 3(a+d) = 33

=> a+d = 11 .

=> d =11-a

Therefore ,Second term of A.P = 11 .

The product of first and third terms = (a)(a+2d) = a(a+2(11-a)

= a(a+22-2a)

= a(22-a)

= 22a-a²

ATQ --->

Given 22a-a²-29= a+d

=> 22a-a²-29=11

=> 22a-a² -40 =0

=> a²-22a+40=0

=> a²-20a-2a+40=0

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


=> a= 2 or 20.

Finding common difference for a = 2
11-2=9

Finding Common difference for a =20
11-20=-9 .

Now The possible A .P 's are
1) 2,11,20,29,38,47,56,65,74.....
2) 20,11,2,-7,-16,-25,-34,-43,-52,-61,-70 .......

Answer 2

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...!!!