Question
The sum of all real values of x satisfying the equation
is:
-
3
-
-4
-
6
-
5
Solution
A.
3
Given, 
Sponsor Area
The sum of all real values of x satisfying the equation is:
3
-4
6
5
A.
3
Given,
Sponsor Area
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:
46th
59th
52nd
58th
D.
58th
Clearly, number of words start with A = 4!/ 2! = 12
Number of words start with L = 4! = 24
A number of words start with M = 4!/2! = 12
Number of words start with SA = 3!/2! = 3
Number of words starts with SL = 3! = 6
Note that, next word will be 'SMALL'
Hence, the position of word 'SMALL' is 58th.
If the number of terms in the expansion of 
is 28, then the sum of the coefficients of all the terms in this expansion is
64
2187
243
729
D.
729
Clearly, number of terms in the expansion of 
If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:
8/5
4/3
1
7/4
B.
4/3
Let a be the first term and d be a common difference. Then, we have a+d, a+4d, a+8d in GP,
ie. (a +4d)2 = (a+d)(a+8d)
⇒ a2 +16d2 +8ad = a2 +8ad +ad +8d2
⇒ 8d2 = ad
8d =a [∴ d≠0]
Now, common ratio,
If the sum of the first ten terms of the series,
is 16/5 m, the m is equal to
102
101
100
99
B.
101
Let S10 be the sum of first ten terms of the series. Then, we have
Sponsor Area
Mock Test Series
