Question

In an ordered set of four numbers, the first three are in

a.p and last three are in g.p whose common ratio is 7/4 if the product of the first and fourth of these numbers is 49, then the product of the second and third of these is

Answer 1

answer is 28? reciprocal of 7/4 = 4/7x49
4x7=27

Answer 2

Answer:  112

Step-by-step explanation:

Let the first number is a,

Since, First three numbers are in A.P.,

Let d be the common difference,

First three numbers are out of four numbers,

a, a+d, a+2d

Now, If last three numbers are in G.P,

First two numbers of G.P are,

a+d, a+2d

⇒ Common ratio = (a+2d)/(a+d)

⇒ Third number of G.P = (a+2d)× (a+2d)/(a+d) = (a+2d)²/(a+d)

Hence, the order of the numbers is,

a, a+d, a+2d, (a+2d)²/(a+d)

Now, According to the question,

$a\times \frac{(a+2d)^2}{a+d} = 49$ ------ (1)

Also,

$\frac{a+2d}{a+d} = \frac{7}{4}$ -----(2)

By equation (2),

4(a+2d) = 7(a+d) ⇒ 4a + 8d = 7a + 7d ⇒ d = 3a

Put this value in Equation (1),

$a\times \frac{ (a+6a)^2}{a+3a} = 49$

$\frac{(7a)^2}{4}=49$

$(7a)^2=4\times 49$

$7a = 2\times 7$

$a=2$

$\implies d = 6$

Hence, the product of the second and third of these is,

(a+d)×(a+2d) = (2+6)(2+12)=8 × 14 = 112