Question

OPEN CHALLENGE : BEST GETS BRAINLIEST : Solve the following questions specifically step by step and correctly :

1.) In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. ( Assumption : The 3 consecutive terms are a-d, a, a+d )

2.) Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14. (Assumption : the 4 consecutive terms are a-d, a, a+d, a+2d )

3.) If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.

Answer 1

Heya friend..
Here is yr answer...
1. Let the numbers in AP be a-d, a, a+d
ATQ,
a-d+a+a+d=27.
3a =27
a=9
Now,
(a-d) (a)(a+d) =504
(a²-d²) a =504
a³-ad²=504
Putting the value of a as 9
9³-9×d²=504
729-9d²=504
-9d²=504-729
d²=225/9
d=5
Required terms are 4,9 and 14....
2. ATQ
a-d+a+a+d+a+2d =12
2a+2d =12
a+d =6
a=6-d --------equation 1
Now,
a+d+a+2d=14
2a+3d=14
Putting the value of a as 6-d
2(6-d)+3d=14
12 - 2d +3d =14
-d=2
d=-2
Then a=8
The required numbers are 10, 8,6,4.....
3. ATQ
a+8d =0
a=-8d
19 th term =a+18 d =-8d+18d=10 d
29 th term =a+28d =-8d+28d=20 d =2(10d)..
Therefore 29th term =2×19th term proved..
Hope this helps u... Plzzz mark as brainliest.

Answer 2

According to Question 
a-d+a+a-d+a+2d=12
2a+2d=12
a+d=6
a=6-d.....eqn(1)
now,
(2) a+d+a+2d+14
2a+3d+14putting the value of a as 6-d
2(6-d)+3d =14
12 - 2d +3d=14
d=2
d = -2
then a= 8 
the required number are 10,8,6,4...
(3) according to question 
a+8d=0 
a=-8
19th term u+18d =-8d +18d =10d
29th term =a +28d =-8d+28d=20d 2(10d)
 therefore 29th term =2*19th term provide