Question

Find the four numbers in A.P such that the sum of 2nd and 3rd term is 22 and the product of 1st and 4th term is 85​

Answer 1

Answer:

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Answer 2

Answer:

hii

here is your answer

Formula of nth term= a_n=a+(n-1)d

Substitute n = 2

a_2=a+(2-1)d

a_2=a+d

Substitute n = 3

a_3=a+(3-1)d

a_3=a+2d

Substitute n = 1

a_1=a

Substitute n = 4

a_4=a+(4-1)d

a_4=a+3d

We are given that the sum of 2nd and 3rd term is 22

So, a+d+a+2d=22

2a+3d=22 --- 1

Now we are given that the product of first and fourth term is 85

So, a(a+3d)=85

a^2+3ad=85

Substitute the value of a from 1

(\frac{22-3d}{2})^2+3(\frac{22-3d}{2})d=85

d=4,-4

Substitute d = 4 in 1

2a+3(4)=22

2a+12=22

2a=10

a=5

So, first term = 5

AP = 5,5+4,5+4+4,5+4+4+4,...

AP = 5,9,13,17,...

Substitute d =- 4 in 1

2a+3(-4)=22

2a-12=22

2a=34

a=17

So, first term =17

AP = 17,17-4,17-4-4,17-4-4-4,...

AP = 17,13, 9 , 5 ..

hope this will help uhh....