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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-
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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.
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| Question Text | 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- |
| Answer Type | Text solution:1 |
| Upvotes | 150 |
