Class 11

Math

Algebra

Sequences and Series

If the sum of four number in $$A.P.$$ be $$48$$ and that the product of the extremes is to the product of the means is $$27$$ to $$35$$ then the number are-

- $$3, 9, 15, 21$$
- $$9, 5, 7, 3$$
- $$6, 10, 14, 18$$
- None of these

Let the terms of AP be $$a-3d, a-d, a+d, a+3d$$

so, $$a-3d+a-d+a+d+a+3d=48$$

$$\Rightarrow 4a=48$$

$$\Rightarrow a=12$$

$$\Rightarrow \dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}=\dfrac{27}{35}$$

$$\Rightarrow \dfrac{a^2-9d^2}{a^2-d^2}=\dfrac{27}{35}$$

$$\Rightarrow 35a^2-315d^2=27a^2-27d^2$$

$$\Rightarrow 8a^2-288d^2=0$$

$$\Rightarrow 8(12)^2-288d^2=0$$

$$\Rightarrow d^2=\dfrac{8\times 12\times 12}{288}=4$$

$$d=\pm 2$$

Case $$1$$ $$\Rightarrow d=2; a=12$$

Terms$$=6, 10, 14, 18$$

Case $$2$$ $$\Rightarrow d=-2, a=12$$

Terms$$=18, 14, 10, 6$$

Option $$[C]$$ is correct.