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Bhakti asks : . Write first four terms of the AP, when the first term a= -3 and the common difference d =0 a is the first term next term is obtained by adding d to the preceding term.  
Category : Quantitative Ability |  On : 2012-03-14 04:44:48
 
Rolly singh says :
14/03/2012 05:50:38
series contains same number -3,-3,-3,-3
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Bhakti asks : Write first four terms of the AP, when the first term a=10 and the common difference d =10rn( a is the first term, succeeding term is obtained by adding d to the preceding term.)rnA) 10, 20, 30, 40rnB) 40, 30,20,10rnC) 10, -20, -30, -40rnD) -40, -30, -20, -10 
Category : Quantitative Ability |  On : 2012-03-14 10:19:19
 
Rk singh says :
14/03/2012 03:09:37
the answer is a) 10, 20, 30, 40. Please use the formula nth term t_n = a + (n - 1)d.
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Harminder singh asks : what is the value of sum of n terms of the series whose nth term is(2sinA) raised to power n where n belongs to +ve integers. help yar, I am really confused for it. I just need it to solve a big problem of my life 
Category : Mathematics |  On : 2012-03-05 09:18:28
 
Rolly singh says :
09/03/2012 09:01:51
2 sinA + (2 sina)^2................ apply the summation formula of gp........a(1-r^n)/r-1
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Lalit shantilal kariya asks : The ratio of sums of n terms of two A.P. 's are in the ratio (7n-5) : (5n+17). Show that the 6th terms of both the progression are same. 
Category : CBSE |  On : 2012-03-05 12:28:39
 
08/03/2012 06:05:09
this is an ncert question of class 11th...mathematics cbse
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Abhijeet singh asks : (121)^x=(0.125)^y=(484)^z,then find the value of z in terms of x and y. a)2xy/(3y-2x) b)3xy/(3y-2x) c)xy/(3y-2x) d)3xy/(2x-3y) 
Tag : zthen,terms
Category : MBA Entrance |  On : 2012-03-01 05:25:04
 
Gaurav says :
22/03/2012 11:17:33
above eq can written as: 11^2x = (5/10)^3y = 22^2z or 11^2x = (2)^(-3y) = 2^2z . 11^2z multiply first two and equate with square of 3rd as all are equal 11^2x . 2^(-3y) = (2^2z . 11^2z ) ^2 now equate LHS & RHS powers which having same base: for power of 11 => 2x = 4z => z= x/2 -------------(a) for power of 2 => -3y = 4z => z= -3y/4 --------------(b) above two x/2 = -3y/4 => 2x = -3y -----------------------------------(c) now check option (hit and trial) try option 2: z = 3xy/(3y-2x) put value of 2x in denominator from eq c ===> z = 3xy / (3y +3y) ==> z= 3xy/6y => z=x/2 which is same as eq a HENCE option B: z = 3xy / (3y-2x)
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