Question

find four numbers in AP such that sum of 2nd and 3rd term is 22 and the product of first and fourth term is 85

Answer 1

Let the numbers be
a-3d, a-d, a+d, a+3d
$(a - d) + (a + d) = 22 \\ a = 11$
$(a - 3d)(a + 3d) = 85 \\ {a}^{2} - 9 {d}^{2} = 85 \\ 9 {d}^{2} = {11}^{2} - 85 = 36 \\ d = \sqrt{4} = 2 \: \: or \: \: - 2$
So the numbers are
5, 9, 13, 17
or
17, 13, 9, 5

Answer 2

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 ..