Question

The sum of four consecutive numbers in an A.P.is
32 and the ratio of the product of the first and the
last term to the product of two middle terms is 7:15
Find the numbers. ​

Answer 1

Answer:

Given:

• Sum of four consecutive terms in an AP is 32.
• And the ratio of the product of the first and the last term to the product of two middle terms is 7:15.

To find:

• The numbers in the series=?

Solution:

Let the four consecutive terms in an AP is,

→ a-3d,a-d,a+d,a+3d

According to the question,

Case-1:-

Sum of four consecutive terms in an AP is 32.

→ a-3d+a-d+a+d+a+3d=32

→ 4a=32

→ a=8

So,The first term,a=8.

case-2:-

The ratio of the product of the first and last term to the product of two middle terms is 7:15.

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15}$

$\frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15}$

$15 {a}^{2} - 135 {d}^{2} = 7 {a}^{2} - 7 {d}^{2}$

$8 {a}^{2} = 128 {d}^{2}$

Since,

The value of a is 8.

$8(64) = 128 {d}^{2}$

${d}^{2} = 4$

$d = 2$

So, The common difference, d=2

Now,

The series is,

→ a-3d=8-6=2

→ a-d=8-2=6

→ a+d=10

→ a+3d=8+6=14

The series of AP is 2,6,10 and 14.

Answer 2

a=8
d=2
I hope this is helpful