JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 21)
If the sum and product of the first three term in
an A.P. are 33 and 1155, respectively, then a value
of its 11th term is :-
–25
–36
25
–35
Explanation
Let the three terms are a - d, a, a + d
Given a - d + a + a + d = 33
$$ \Rightarrow $$ 3a = 33
$$ \Rightarrow $$ a = 11
Also given,
(a - d)a(a + d) = 1155
$$ \Rightarrow $$ (a2 - d2)a = 1155
$$ \Rightarrow $$ (112 - d2)11 = 1155
$$ \Rightarrow $$ (112 - d2) = 105
$$ \Rightarrow $$ d = $$ \pm $$ 4
When d = 4 and a = 11 then series is
7, 11, 15, ....
$$ \therefore $$ T11 = a + 10d = 7 + 10$$ \times $$ 4 = 47
When d = -4 and a = 11 then series is
15, 11, 7, ....
$$ \therefore $$ T11 = a + 10d = 15 + 10$$ \times $$ -4 = -25
Given a - d + a + a + d = 33
$$ \Rightarrow $$ 3a = 33
$$ \Rightarrow $$ a = 11
Also given,
(a - d)a(a + d) = 1155
$$ \Rightarrow $$ (a2 - d2)a = 1155
$$ \Rightarrow $$ (112 - d2)11 = 1155
$$ \Rightarrow $$ (112 - d2) = 105
$$ \Rightarrow $$ d = $$ \pm $$ 4
When d = 4 and a = 11 then series is
7, 11, 15, ....
$$ \therefore $$ T11 = a + 10d = 7 + 10$$ \times $$ 4 = 47
When d = -4 and a = 11 then series is
15, 11, 7, ....
$$ \therefore $$ T11 = a + 10d = 15 + 10$$ \times $$ -4 = -25
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