For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... are equal?
Answers
Answer:
The value of n is 13
Step-by-step explanation:
Given :
Case : 1
Let the first term of this AP be 'A' , the common difference be 'D' and the nth term be 'An'.
A= 63, D = 65 - 63 = 2
Case : 2
Let the first term of this AP be 'a' , the common difference be 'd' and the nth term be 'an'
a = 3, d = 10 - 3 = 7
An = an [Given]
A + (n -1)D = a + (n -1)d
63 + (n -1)(2) = 3 + (n -1)(7)
63 + 2n - 2 = 3 + 7n - 7
61 + 2n = - 4 + 7n
2n - 7n = - 4 - 61
-5n = - 65
n = - 65/-5
n = 13
Hence, the value of n is 13
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Answer:
Given
Case 1st
AP sequence
63, 65 , 67 ......
A = 63
Common difference ( d ) = a² - a¹
d = 65 -63
d = 2
Now , Formula for an is
an = a + (n-1) d
an. = 63 + (n-1) 2
an. = 63 +2n -2
an. = 61 + 2n
Case 2nd
AP sequence
3, 10 ,17......
a = 3
d = 10 -3
d = 7
an = a+(n-1)d
an = 3 +(n-1)7
an = 3+ 7n -7
an = 7n -4
As given in the question
Value of nth term is equal ,so
61 + 2n = 7n -4
61 +4 = 7n -2n
65. = 5n
65/5 = n
13. = n
Value of n is 13
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