Question

4 numbers are in ap in which the product of 1st & last term is 16 & sum of middle 2 terms is 10 find the numbers

Answer 1

Answer:

2,4,6,8 or 8,4,6,2

Step-by-step explanation:

Let first term of AP be = a

Common difference be = d

We know that nth term of an AP is given by

a(n) = a + (n-1)d

Therefore, fourth term will be

a(4) = a + (4-1)d

a(4) = a + 3d

Similarly

a(3) = a + 2d

a(2) = a + d

Now,

a(1) × a(4) = 16

a × { a + 3d } = 16 ...........[1]and

a(2) + a(3) = 10

(a+d) + (a+2d) = 10

(2a + 3d) = 10

3d = 10-2a ...........[2]

Putting [2] in [1]

a × (a + 10-2a) = 16

On solving we get

a = 2 or 8

for a = 2 d = 2

and for a = 8 d = -2

Therefore possible APs are

2,4,6,8 and 8,6,4,2