Question

the sum of three consecutive terms of an ap is 3 and thier product is -35. find the numbers

Answer 1

Answer:

Step-by-step explanation:

By solving with the help of the equation

Sn=n/2(a+l)

We get n=10

Puting the value in

Tn=a+(n-1)d

By solving this equation

We get d=3

HOPE THIS WILL HELP U

Answer 2

Hey!
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★Arithmetic Progressions ★
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=> Let the three consecutive terms be ( a - d ) , a and ( a + d )


➡According to the Question ,


• There Sum is 3 so we will add the terms of A.P


=> a + a + d + a - d = 3

=> 3a = 3

=> a = 1 ✔


•°• The terms become ( 1 - d ) , 1 and ( 1 + d)


•their product is -35 . So multiply the terms.


=> 1 ( 1 + d ) ( 1 - d ) = -35

=> 1 - d² = -35

=> d² = 36

=> d = +- 6 ✔


Now we shall take both negative as well as positive value of d.


1. When d = 6

★The terms of the AP are -5 , 1 and 7 !


2. When d = -6

★The terms of the A.P are 7 , 1 and -5 !



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