Q1.tell the AP whose 23rd term is 93 and the 32nd term is 129.
Answers
Answer:
Given
23rd term is 93 & 32nd term is 129.
To Find
We have to find the AP
Solution:
23rd term can be written as : a + (23-1)d
=> a + 22d=93 ------(1)
where ,a is the first term and d is the common difference
from equation 1
=> a = 93-22d ------(2)
32nd term can be written as: a+(32-1)d
=>a+31d= 129 -----(3)
Put Equation 2 into Equation 3
➙93-22d+31d= 129
➙93+9d= 129
➙9d= 129-93
➙9d= 36
➙d= 36÷9
➙d= 4
Therefore,the common difference is 4.
Put d into Equation 2
➙a= 93-22d
➙a= 93-22(4)
➙a= 93-88= 5
Hence, first term is 5 .
Our AP will be a,a+d,a+2d,a+3d
➙5 , 5+4, 5+2(4),5+3(4)
➙5,9,13,17
AP is 5,9,13,17..
Answer:
Given
23rd term is 93 & 32nd term is 129.
To Find
We have to find the AP
\sf\huge\bold{\underline{\underline{{Solution}}}}
Solution
23rd term can be written as : a + (23-1)d
=> a + 22d=93 ------(1)
where ,a is the first term and d is the common difference
from equation 1
=> a = 93-22d ------(2)
32nd term can be written as: a+(32-1)d
=>a+31d= 129 -----(3)
Put Equation 2 into Equation 3
➙93-22d+31d= 129
➙93+9d= 129
➙9d= 129-93
➙9d= 36
➙d= 36÷9
➙d= 4
Therefore,the common difference is 4.
Put d into Equation 2
➙a= 93-22d
➙a= 93-22(4)
➙a= 93-88= 5
Hence, first term is 5 .
Our AP will be a,a+d,a+2d,a+3d
➙5 , 5+4, 5+2(4),5+3(4)
➙5,9,13,17
AP is 5,9,13,17..