Question

the sum of four numbers in ap is 28 and product of 1st and 4th is to the product of 2nd and 3rd is 5:6.find the numbers

Answer 1

let the for Numbers be (a-3d), (a-d), (a+d) and (a+3d),

then
according to the question, we have

(a-3d)+(a-d)+(a+d)+(a+3d)=28,

then

4a=28,

then

a=7,


also

(a-3d)(a+3d)/(a-d)(a+d) = 5/6,

6(a²-9d²)=5(a²-d²) ,

6a²-54d²=5a²-5d²,

then

6a²-5a²=-5d²+54d²,

a²=49d²,

7²=49d²,

49=49d²,

then

d²=1,

hence

d=1,


therefore

1st number=(7-3×1)=7-3=4,

2nd number=(7-1)=6,

3rd number=(7+1)=8,

4th number=(7+3×1)=7+3=10

3

Answer 2

$\huge\green{\mathfrak{===================}}$
$\huge\pink{\mathfrak{Bonjour!!}}$

$\green{\mathfrak{given}}$ :sum of four no. is 28 and

$\frac{t1 \times t4}{t2 \times t3} = \frac{5}{6}$

now,

(a-3d)+(a-d)+(a+d)+(a+3d)=28

4a=28

a=7

$\green{\mathfrak{we \:have: }}$
a-3d)(a+3d)/(a-d)(a+d) = 5/6,

6(a²-9d²)=5(a²-d²) ,

6a²-54d²=5a²-5d²,

then

6a²-5a²=-5d²+54d²,

a²=49d²,

7²=49d²,

49=49d²,

then

d²=1,

hence d=1

Ap are:$\huge{\boxed{a-3d=4\:, \: a-d=6 \: ,a+d=8,a+3d=10}}$