Solve previous year Question of iit jee
Chapter :- sequence and series
Answers
Answer:
Option d
Step-by-step explanation:
Given :-
100 times the 100th term of an AP with non-zero Common difference equal to the 50 times it's 50th term.
To find :-
Find the 150th term of the AP ?
Solution :-
We know that
The general term of an AP = an = a+(n-1)d
Where, a = First term
d = Common difference
n = Number of terms
Now , 50th term = a 50
=> a 50 = a+(50-1)d
=> a 50 = a +49d
100th term = a 100
=> a 100 = a +(100-1)d
=> a 100 = a +99d
Given that
100 times the 100th term of an AP with non-zero Common difference equal to the 50 times it's 50th term.
=> 100× a100 = 50× a50
=> 100×(a+99d) = 50×(a+49d)
=> 100/50 × (a+99d) = a+49d
=> 2(a+99d) = a+49d
=> 2a+198d = a+49d
=> 2a+198d -a - 49d = 0
=> (2a-a) +(198d-49d) = 0
=> a +149d = 0
=> a +(150-1)d = 0
=> a150 = 0
Therefore, a150 = 0
Answer:-
The 150th term of the given AP is Zero for the given problem.
Used formulae:-
The general term of an AP = an = a+(n-1)d
Where, a = First term
d = Common difference
n = Number of terms
Answer:
Option (D) , T₁₅₀ = 0 is correct option
Step-by-step explanation:
Given :- 100times the 100th term of an AP = 50 times of its 50th term
to find The 150th term of the same AP
Let a be the first term and d (d≠0) be the common difference of the given AP , then
- T₁₀₀ = a + 99d ___(1)
- T₅₀ = a + 49d _____(2)
- T₁₅₀ = a+149d ______(3)
Now, According to the given data ,
- 100× T₁₀₀ = 50× T₅₀
- => 100(a+99d) = 50 × (a+ 49d)
- => 2(a+99d) = a+49d
- => a + 149d = 0
- From eq (iii) T₁₅₀ = 0 Answer