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Q.1. Let S_{k}, k = 1, 2, ….. , 100, denote the sum of the infinite geometric series whose first term is
and the common ratio is . Then the value of
is (2010)
Ans. (3)
Sol. Using we get
Q.2. Let a_{1},a_{2},a_{3}........, a_{11 }be real numbers satisfying
a_{1}=15, 27–2a_{2}> 0 and a_{k}=2a_{k–1}–a_{k–2} for k = 3, 4,..........11.
if , then the value of
is equal to (2010)
Ans. (0)
Sol. Given that a_{k} = 2a_{k –1}–a_{k –2}
⇒ a_{1 }, a_{2} , a_{3} , ...,a_{11} are in AP..
If a is the first term and D the commen difference then
a^{2}_{1} + a^{2}_{2} + ... +a^{2}_{11}= 990 (
⇒ 11a ^{2} + d ^{2} (1^{2} + 2^{2} + ...+ 10^{2} ) + 2ad (1 + 2 + ...+ 10) = 990
⇒ a^{2 }+ 35d^{2} + 150d= 90 Using a = 15,
we get 35d^{2} + 150d +135 = 0 or 7d^{2} + 30d + 27= 0
⇒ d +3)(7d +9) = 0 ⇒ d= –3 or – 9/7
then a^{2} = 15  3 = 12 or 15
∴ d ≠ – 97
Hence
Q.3. Let a_{1}, a_{2}, a_{3} .....a100 be an arithmetic progression with a_{1 }= 3 and For any integer n with 1 ≤ n ≤ 20 , let m = 5n. If does not depend on n, then a_{2} is (2011)
Ans. (9)
Sol. We have
which will be independent of n if d = 6 or d = 0 For a proper A.P. we take d = 6 then a_{2} = 3 + 6 = 9
Q.4. A pack con tain s n cards n umber ed fr om 1 to n . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k – 20 = (JEE Adv. 2013)
Ans. (5)
Sol. Let k, k + 1 be removed from pack.
∴ (1 + 2 + 3 + ... + n) – (k + k + 1) = 1224
for n = 50, k = 25 ∴ k – 20 = 5
Q.5. Let a, b, c be positive integers such that is an integer. If a,b, c are in geometric progression and the arithmetic mean of a, b, c is b + 2, then the value of is
(JEE Adv. 2014)
Ans. (4)
Sol. ∵ a, b, c are in G.P
∴ b = ar and c = ar^{2}
Also is an integer
⇒ r is an integer
∵ A.M. of a, b, c is b + 2
⇒ a + ar + ar^{2} = 3ar+6
⇒ a (r^{2 } 2r + 1)=6
⇒ a (r  1)^{2}=6 .
∵ a and r are integers
∴ The only possible values of a and r can be 6 and 2 respectively.
Then
Q.6. Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is (JEE Adv. 2015)
Ans. (9)
Sol.
a_{7} = a + 6d = 15d
∵ 130 < 15d < 140 ⇒ d = 9
(∵ All terms are natural numbers ∴ d ∈N )
Q.7. The coefficient of x^{9} in the expansion of (1 + x) (1 + x^{2}) (1 + x^{3}) ... (1 + x^{100}) is (JEE Adv. 2015)
Ans. (8)
Sol. In expansion of (1 + x) (1 + x^{2}) (1 + x^{3}) .... (1 + x^{100}) x^{9 }can be found in the following ways x^{9}, x^{1 +2} 8, x ^{2 + 7}, x^{3 + 6}, x^{4 + 5}, x^{1 + 2 + 6}, x^{1} ^{+ 3 + 5}, x^{2 + 3 + 4 }The coefficient of x9 in each of the above 8 cases is 1. ∴ Required coefficient = 8.
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