The 5th and 15th term of an A.p are 13and 17find that A.P
Answers
S O L U T I O N : (ques.error)
The 5th & 15th term of an A.P are 13 & -17.
As we know that formula of an A.P;
- a is the first term.
- d is the common difference.
- n is the term of an A.P.
A/q
➠ a5 = 13
➠ a + (5 - 1)d = 13
➠ a + 4d = 13...............(1)
&
➠ a15 = -17
➠ a + (15 - 1)d = -17
➠ a + 14d = -17.........…...(2)
Now, using by substitution method :
From equation (1), we get;
➠ a + 4d = 13
➠ a = 13 - 4d.............(3)
Therefore, putting the value of a in equation (2),we get;
➠ 13 - 4d + 14d = -17
➠ 13 + 10d = -17
➠ 10d = -17 - 13
➠ 10d = -30
➠ d = -30/10
➠ d = -3
Therefore, putting the value of d in equation (3),we get;
➠ a = 13 - 4 × (-3)
➠ a = 13 - (-12)
➠ a = 13 + 12
➠ a = 25
Now,
The arithmetic progression :
- a = 25
- a + d = 25 + (-3) = 25 - 3 = 22
- a + 2d = 25 + 2(-3) = 25 +(-6) = 25 - 6 = 19
- a + 3d = 25 + 3(-3) = 25 + (-9) = 25 - 9 = 16
Thus,
The AP will be 25, 22, 19, 16....
Answer:
S O L U T I O N : (ques.error)
Given:
The 5th & 15th term of an A.P are 13 & -17.
Explantion:
As we know that formula of an A.P;
a
n
=a+(n−1)d
a is the first term.
d is the common difference.
n is the term of an A.P.
A/q
➠ a5 = 13
➠ a + (5 - 1)d = 13
➠ a + 4d = 13...............(1)
&
➠ a15 = -17
➠ a + (15 - 1)d = -17
➠ a + 14d = -17.........…...(2)
Now, using by substitution method :
From equation (1), we get;
➠ a + 4d = 13
➠ a = 13 - 4d.............(3)
Therefore, putting the value of a in equation (2),we get;
➠ 13 - 4d + 14d = -17
➠ 13 + 10d = -17
➠ 10d = -17 - 13
➠ 10d = -30
➠ d = -30/10
➠ d = -3
Therefore, putting the value of d in equation (3),we get;
➠ a = 13 - 4 × (-3)
➠ a = 13 - (-12)
➠ a = 13 + 12
➠ a = 25
Now,
The arithmetic progression :
a = 25
a + d = 25 + (-3) = 25 - 3 = 22
a + 2d = 25 + 2(-3) = 25 +(-6) = 25 - 6 = 19
a + 3d = 25 + 3(-3) = 25 + (-9) = 25 - 9 = 16
Thus,
The AP will be 25, 22, 19, 16....