10th term of an arithmetic sequence is 64. If its common difference is 7, find the position of the term 400 in the sequence?
Answers
Step-by-step explanation:
\underline{\textbf{Given:}}
Given:
\textsf{10 th term of an arithmetic sequence is 64}10 th term of an arithmetic sequence is 64
\textsf{and common differene is 7}and common differene is 7
\underline{\textbf{To find:}}
To find:
\textsf{The position of the term 400}The position of the term 400
\underline{\textbf{Solution:}}
Solution:
\underline{\textbf{Concept used:}}
Concept used:
\textsf{The n th term of the A.P a, a+d, a+2d, . . . . . . . . is }The n th term of the A.P a, a+d, a+2d, . . . . . . . . is
\boxed{\mathsf{t_n=a+(n-1)d}}
t
n
=a+(n−1)d
\mathsf{Consider,}Consider,
\mathsf{t_{10}=64}t
10
=64
\implies\mathsf{a+9d=64}⟹a+9d=64
\implies\mathsf{a+9(7)=64}⟹a+9(7)=64
\implies\mathsf{a+63=64}⟹a+63=64
\implies\mathsf{a+63=64}⟹a+63=64
\implies\textbf{a=1}⟹a=1
\mathsf{Now,}Now,
\mathsf{t_n=400}t
n
=400
\mathsf{a+(n-1)d=400}a+(n−1)d=400
\mathsf{1+(n-1)7=400}1+(n−1)7=400
\mathsf{(n-1)7=400-1}(n−1)7=400−1
\mathsf{(n-1)7=399}(n−1)7=399
\mathsf{n-1=\dfrac{399}{7}}n−1=
7
399
\mathsf{n-1=57}n−1=57
\mathsf{n=57+1}n=57+1
\implies\boxed{\mathsf{n=58}}⟹
n=58
\therefore\textbf{58 th term is 400}∴58 th term is 400
Given: 10th term of an arithmetic sequence = 64
Common difference of the sequence = 7
To find: The position of the term 400 in the sequence
Solution: The nth term of an arithmetic progression is given by the formula :-
an = a + (n - 1)d [where a is the first term and d is the common difference]
Now according to the question,
a₁₀ = 64
⇒ a + (10 - 1)7 = 64
⇒ a + 9 × 7 = 64
⇒ a = 64 - 63
⇒ a = 1
Since the values of both a and d are known to us now, we can calculate the position of the term 400 in the sequence.
Accordingly,
a + (n - 1)d = 400
⇒ 1 + (n - 1)7 = 400
⇒ (n - 1)7 = 400 - 1
⇒ (n - 1)7 = 399
⇒ n - 1 = 399/7
⇒ n = 57 + 1
⇒ n = 58
Therefore, the position of the term 400 in the sequence is 58.