find the 41st term of an ap whose 11th term is 37 and the 16 th term is 52
Answers
Answer:
127
Step-by-step explanation:
11th term = a + 10d = 37
16th term = a + 15d = 52
by, solving, d = 3 , a = 7
41 st Term = a + 40 d
= 7 + 40 × 3
= 127
Answer :-
- 41st term of AP is 127
Step-by-step explanation:
Given : 11th term of AP is 37 and 16th term of AP is 52
To find : 41st term of AP
Solution :
In order to find the 41st term of AP, we have to find the value of common difference and first term of AP.
Given that the 11th term of AP is 37, which can be expressed as :
⟶ a + 10d = 37 - - - Eqn(1)
Here :-
- a = first term
- d = common difference
Similarly 16th term can be expressed as:
⟶ a + 15d = 52 - - - Eqn(2)
Now subtract equation 1 from equation 2 :-
Eqn (2) - Eqn(1)
↪ a + 15d - (a + 10d) = 52 - 37
↪ a + 15d - a - 10d = 15
↪ a - a + 15d - 10d = 15
↪ 5d = 15
↪ d = 15/5
↪ d = 3
Now substitute this value of d in Eqn (1) or (2).
I am substituting it in equation (1) you can substitute it in any of the two.
⟹ a + 10d = 37
⟹ a + 10 (3) = 37
⟹ a + 30 = 37
⟹ a = 37 - 30
⟹ a = 7
Now this question seems to be an easy one !
We have obtained values of first term and common difference, now we have to find the 41st term.
For obtaining the required term, we are applying formula for obtaining nth term.
→ an = a + (n - 1)d
Here :-
- an = nth term
- a = first term
- d = common difference
Substituting the values of a and d
→ an = 7 + ( n - 1 ) ( 3 )
Substitute n = 41, for 41st term
→ a41 = 7 + ( 41 - 1 ) ( 3 )
→ a41 = 7 + ( 40 ) ( 3 )
→ a41 = 7 + 120
→ a41 = 127
So the required term (41st) of the AP is 127.