CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The sum of four consecutive numbers 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 $$7:15$$. Find the numbers.


Solution

Let,the numbers in A.P be $$a-3d,a-d,a+d,a+3d$$

Given: Sum of four consecutive numbers $$=32$$

$$\therefore a-3d+a-d+a+d+a+3d=32$$

$$\implies 4a=32$$

$$\implies a=\dfrac{32}{4}=8$$

Now,

A.T.Q,

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

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

$$\implies \dfrac{64-9d^2}{64-d^2}=\dfrac{7}{15}[\because a=8]$$

$$\implies (64-9d^2)\times 15=7(64-d^2)$$

$$\implies 64\times 15-135d^2=7\times 64-7d^2$$

$$\implies 64\times 15-7\times 64=135d^2-7d^2$$

$$\implies 64(15-7)=128d^2$$

$$\implies 64\times8=128d^2$$

$$\implies \dfrac{64 \times8}{128}=d^2$$

$$\implies d^2=5$$

$$\implies d=\pm2$$

$$\therefore$$ Either , the numbers are $$2,6,10,14$$

or

$$14,10,6,2$$


Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image