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If 1, log_{9} (3^{1–x} + 2), log_{3} (4.3^{x} – 1) are in A.P. then x equals [2002]
(b) 1, log_{9} (3^{1 – x }+ 2), log_{3 }(4.3^{x – 1}) are in A.P.
⇒ 2 log_{9} (3^{1– x}+2) = 1 + log3 (4.3^{x – 1}) ⇒ log_{3} (3^{1 – x} + 2) = log_{3}3 + log_{3} (4.3^{x – 1})
⇒ log_{3} (3^{1– x} + 2) = log3 [3(4 × 3x – 1)] ⇒ 3^{1– x }+ 2 = 3 (4.3^{x – 1}) ⇒ 3.3^{–x} + 2 = 12.3^{x} – 3.
Put 3^{x} = t
=
Hence or x = log_{3}3 – log3 4
⇒ x = 1 – log_{3}4
l, m, n are the pth, qth and rth term of a G. P. all positive, then equals[2002]
l = AR^{p –1} ⇒ log l = log A + (p – 1) log R
m = AR^{q –1} ⇒ log m = log A + (q – 1) log R
n = AR^{r –1 }⇒ log n = log A + (r – 1) log R
Now,
Operating C_{1}  ( log R ) C_{2} + ( log R logA) C_{3}
The value of 2^{1/4}. 4^{1/8}. 8^{1/16} ... ∞ is [2002]
The product is P = 2^{1/4}.2^{2/8}.2^{3/16}......... = 2^{1/4} ^{+ 2/8 + 3/16+..........}∞
Now let ......(1)
......(2)
Subtracting (2) from (1)
or
∴ P = 2^{S }= 2
4. Fifth term of a GP is 2, then the product of its 9 terms is [2002]
ar^{4} = 2
a x ar x ar^{2}x ar^{3 }x ar^{4}x ar^{5} x ar^{6} x ar^{7}x ar^{8}
= a^{9} r^{36} = (ar^{4})^{9} = 2^{9} = 512
Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is [2002]
Let a = first term of G.P. and r = common ratio of G.P.; Then G.P. is a, ar, ar^{2}
Given
Also
a^{2} = 100(1 – r)(1 + r)... (ii)
From (i), a^{2} = 400(1 – r)^{2};
From (ii), we get 100(1 – r)(1 + r) = 400(1 – r)^{2 }
⇒ 1 + r = 4 – 4r
⇒ 5r = 3 ⇒ r = 3/5.
1^{3} – 2^{3} + 3^{3} – 4^{3 }+...+9^{3 } = [2002]
1^{3} – 2^{3} + 3^{3} – 4^{3} + ........+ 9^{3}
= 1^{3} + 2^{3} + 3^{3} +.......+ 9^{3} – 2(2^{3} + 4^{3} + 6^{3} + 8^{3})
= 2025 – 1600 = 425
The sum of the series [2003]
up to ∞ is equal to
Let T_{r} be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers
and , then a – d equals [2004]
............(1)
..............(2)
(1)  (2) ⇒ (mn)d =
From (1) a =
The sum of the first n terms of the series
is when n is even. When n is odd the sum is [2004]
If n is odd, the required sum is
[∵ (n–1) is even
∴ using given formula for the sum of (n–1)terms.]
If the coefficients of r th, (r + 1)th, and (r + 2)th terms in the the binomial expansion of (1 + y)^{m }are in A.P., then m and r satisfy the equation [2005]
Given ^{m}C_{r 1 }, ^{m}C_{r} , ^{m }C_{r +1} are in A.P..
2^{m} C_{r }= m C_{r 1} + ^{m}C_{r+1}
⇒ m^{2}  m(4r + 1) + 4r^{2}  2=0 .
The sum of series is [2004]
We know that
and
IF where a, b, c are in AA..PP and a  < 1,  b  < 1,  c  < 1 then x, y, z ar e in [2005]
a, b, c are in A.P. OR 2b = a + c
x, y, z are in H.P..
The sum of the series [2005]
.................ad inf. is
Putting x = we get
Let a_{1},a_{2} ,a_{3} ............ be terms on A.P. If then equals [2006]
For
If a_{1} , a_{2} , ........,a_{n} are in H.P., then the expression a_{1}a_{2} + a_{2} a_{3} + ..........+ a_{n 1}a_{n} is equal to [2006]
Also,
Which is the required result.
The sum of series ... upto infinity is [2007]
We know that e^{x} =
Put x = – 1
In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals [2007]
Let the series a, ar, ar^{2}, ..... are in geometric progression. given, a = ar + ar^{2}
[∵ terms of G.P. are positive
∴ r should be positive]
The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is [2008]
As per question,
a + ar = 12 …(1)
ar^{2} + ar^{3 }= 48 …(2)
(Qterms are = + ve and –ve alternately)
⇒ a = –12
The sum to infinite term of the series [2009]
We have
....(1)
Multiplying both sides by we get
....(2)
Subtracting eqn. (2) from eqn. (1) we get
A person is to count 4500 currency notes. Let a_{n} denote the number of notes he counts in the n^{th} minute. If a_{1} = a_{2 }= ... = a_{10} = 150 and a_{10}, a_{11}, ... are in an AP with common difference –2, then the time taken by him to count all notes is [2010]
Till 10^{th} minute number of counted notes = 1500
n^{2}  149n + 3000= 0
⇒ n = 125, 24
But n = 125 is not possible
∴ total time = 24 + 10 = 34 minutes.
A man saves 200 in each of the first three months of his service. In each of the subsequent months his saving increases by 40 more than the saving of immediately previous month. His total saving from the start of service will be 11040 after [2011]
Let required number of months = n
∴ 200 × 3 + (240 + 280 + 320 + ... + (n – 3)^{th} term) = 11040
⇒ (n  3)[240 + 20n  80]= 10440
⇒ (n  3)(20n + 160)= 10440
⇒ (n  3)(n + 8)= 522
⇒ n^{2 }+ 5n  546=0
⇒ (n + 26) (n – 21) = 0
∴ n = 21
Statement1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + .... + (361 + 380 + 400) is 8000
Statement2: for any natural number n. [2012]
nth term of the given series
= T_{n }= (n  1)^{2} +( n 1) n+n^{2}
⇒ n = 20 which is a natural number.
Now, put n =1,2,3,....20
Hence, both the given statements are true and statement 2 supports statement 1.
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,....., is [JEE M 2013]
Given sequence can be written as
Multiply and divide by 9
If (10)^{9} + 2(11)^{1 }(10^{8}) + 3(11)^{2 }(10)^{7 }+ K........ +10 (11)^{9 }= k (10)^{9}, then k is equal to: [JEE M 2014]
Let 10^{9} + 2. (11)(10)^{8} + 3(11)^{2} (10)^{7} + ... + 10(11)^{9} = k(10)^{9 }
Let x = 10^{9} + 2.(11)(10)^{8} + 3(11)2(10)^{7} + ... + 10(11)^{9}
Multiplied by on both the sides
⇒ x = 10^{11} = k.10^{9} Given ⇒ k = 100
Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is: [JEE M 2014]
Let a, ar, ar^{2 }are in G.P.
According to the question a, 2ar, ar^{2} are in A.P.
⇒ 2 × 2ar = a + ar^{2 }
⇒ 4r = 1 + r^{2} ⇒ r^{2 }– 4r + 1 = 0
Since r > 1
is rejected
Hence,
The sum of first 9 terms of the series. [JEE M 2015]
nth term of series =
Sum of n term =
Sum of 9 terms
If m is the A.M. of two distinct real numbers l and n(l, n > 1) and G_{1}, G_{2} and G_{3} are three geometric means between l and n, then equals : [JEE M 2015]
and common ratio of G.P. = r =
= ln (l + n)^{2}
= ln × 2m^{2 }
= 4lm^{2}n
If the 2^{nd}, 5^{th} and 9^{th} terms of a nonconstant A.P. are in G.P., then the common ratio of this G.P. is : [JEE M 2016]
Let the GP be a, ar and ar^{2} then a = A + d; ar = A + 4d; ar^{2} = A + 8d
If th e sum of th e fir st ten ter ms of th e series
then m is equal to : [JEE M 2016]
⇒ m= 101.
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