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

**Solution:**

Given, the sum of four consecutive numbers in an AP is 32.

Ratio of the product of the first and the last terms to the product of the two middle terms is 7:15.

We have to find the number.

Let the consecutive numbers in AP be a - 3d, a - d, a + d, a + 3d.

So, a - 3d + a - d + a + d + a + 3d = 32

4a = 32

a = 32/4

a = 8

Product of the first and last terms = (a - 3d)(a + 3d)

Put a = 8,

= (8 - 3d)(8 + 3d)

= 64 - 9d²

Product of the two middle terms = (a - d)(a + d) = (a² - d²)

Put a = 8,

= 64 - d²

(64 - 9d²)/(64 - d²) = 7/15

15(64 - 9d²) = 7(64 - d²)

15(64) - 135d² = 7(64) - 7d²

15(64) - 7(64) = 135d² - 7d²

64(15 - 7) = (135 - 7)d²

64(8) = 128d²

d² = 64(8)/128

d² = 8/2

d² = 4

Taking square root,

d = ±2

Take d = 2,

When a = 8 and d = 2,

The four consecutive numbers are

a - 3d = 8 - 3(2) = 8 - 6 = 2

a - d = 8 - 2 = 6

a + d = 8 + 2 = 10

a + 3d = 8 + 3(2) = 8 + 6 = 14

Therefore, the four consecutive terms of the AP are 2, 6, 10, 14.

**✦ Try This:** The sum of four consecutive numbers in an AP is 24 and the ratio of the product of the first and the last terms to the product of the two middle terms is 3 : 5. Find the numbers

**☛ Also Check:** NCERT Solutions for Class 10 Maths Chapter 5

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**NCERT Exemplar Class 10 Maths Exercise 5.4 Sample Problem 1**

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

**Summary:**

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15. The numbers are 2, 6, 10 and 14

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