Progression (Sequence and Series)Hard
Question
For a positive integer n let a(n) = 1 + 
, then
, thenOptions
A.a(100) ≤ 100
B.a(100) > 100
C.a(200) ≤ 100
D.a(200) > 100
Solution
Given, a(n) = 1 +
Thus, a (100) < 100,
Therefore, (a) is the answer
Again, a(n)
Therefore, a (200) >
> 100
Therefore, (d) is also the answer.
Thus, a (100) < 100,
Therefore, (a) is the answer
Again, a(n)
Therefore, a (200) >
> 100 Therefore, (d) is also the answer.
Create a free account to view solution
View Solution FreeMore Progression (Sequence and Series) Questions
The common difference of the A.P.: $a_{1},a_{2},\ldots,a_{m}$ is 13 more than the common difference of the A.P.: $b_{1},...The maximum sum of the series 20 + 19 + 18+..... is -...The sum of the infinite series 12 + 22 x + 32 x2 + ..... is-...If the roots of the equation (b − c) x2 + (c − a)x + (a − b) = 0 are equal , then a, b, c will be in-...If nth term of an AP is 1/3 ( 2n + 1) , then the sum of its 19 terms is-...