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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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Divide 20 into four parts which are in A.P. and such that the product of the first and fourth is to the product of the second and third in the ratio 2:3.
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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 |
