JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 12)
Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -$${1 \over 2}$$ , then the greatest number amongst them is:
$${{21} \over 2}$$
27
7
16
Explanation
Let the A.P is
a - 2d, a - d, a, a + d, a + 2d
$$ \because $$ sum = 25
$$ \Rightarrow $$ 5a = 25 $$ \Rightarrow $$ a = 5
Also given,
product (a2 – 4d2) (a2 – d2).a = 2520
$$ \Rightarrow $$ (25 – 4d2) (25 –d2)5 = 2520
$$ \Rightarrow $$ 4d4 – 121d2 – 4d2 + 121 = 0
$$ \Rightarrow $$ (d2 – 1) (4d2 – 121) = 0
$$ \Rightarrow $$ d = $$ \pm $$1, d = $$ \pm {{11} \over 2}$$
When d = $$ \pm $$1 we can't get any fraction term like -$${1 \over 2}$$.
$$ \therefore $$ d = $$ \pm {{11} \over 2}$$
And when d = $${{11} \over 2}$$
we get largest term = 5 + 2d = 5 + 11 = 16
a - 2d, a - d, a, a + d, a + 2d
$$ \because $$ sum = 25
$$ \Rightarrow $$ 5a = 25 $$ \Rightarrow $$ a = 5
Also given,
product (a2 – 4d2) (a2 – d2).a = 2520
$$ \Rightarrow $$ (25 – 4d2) (25 –d2)5 = 2520
$$ \Rightarrow $$ 4d4 – 121d2 – 4d2 + 121 = 0
$$ \Rightarrow $$ (d2 – 1) (4d2 – 121) = 0
$$ \Rightarrow $$ d = $$ \pm $$1, d = $$ \pm {{11} \over 2}$$
When d = $$ \pm $$1 we can't get any fraction term like -$${1 \over 2}$$.
$$ \therefore $$ d = $$ \pm {{11} \over 2}$$
And when d = $${{11} \over 2}$$
we get largest term = 5 + 2d = 5 + 11 = 16
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