Question

Answer the following questions.
34. The sum of four consecutive terms of an AP is 32 and the ratio of the product of the extremes
to the product of their middle terms is 7:15. Find the numbers.​

Answer 1

Given :-

The sum of 4 consecutive number is 32

The product of extreme terms and The product of middle terms are given in ratio 7:15

Solution :-

Let the 4 consecutive numbers be

a - 3d , a - d, a + d, a + 3d

The sum of 4 consecutive number is 32

Therefore,

a - 3d + a - d + a + d + a + 3d = 32

4a = 32

a = 8

The ratio of the product of the extreme terms to the product of the middle terms is 7 :15

Therefore,

( a - 3d) ( a + 3d) / ( a - d) ( a + d) = 7/15

a^2 - 9d^2 / a^2 - d^2 = 7/15

By cross multiply,

15( a^2 - 9d^2 ) = 7( a^2 - d^2 )

15a^2 - 135d^2 = 7a^2 - 7d^2

8a^2 = 128d^2

As we know the value of a = 8

Therefore,

8( 8)^2 = 128d^2

512 = 128d^2

d^2 = 512/128

d^2 = 4

d = 2

so, 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

Hence, The four consecutive terms of an AP are 2 , 6 , 10 , 14 $.$

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━━━━━Let The 4 consecutive numbers in AP be

(a−3d),(a−d),(a+d),(a+3d)

According to the question,

$a−3d+a−d+a+d+a+3d=32$

$4a=32$

$a=8$

Now,

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

15(a²– 9d²) = 7(a²– d²)

15a²– 135d²= 7a² – 7d²

8a²=128d²

Putting the value of a, we get,

d²=4

d = ±2

So, the four consecutive numbers are 2,6, 10, 14

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