Question

Determine the AP whose 23rd term is 93 and the 32nd term is 129​

Answer 1

Given

23rd term is 93 & 32nd term is 129.

To Find

We have to find the AP

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

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 2

Step-by-step explanation:

Given

23rd term is 93 & 32nd term is 129.

To Find

We have to find the AP

$\sf\huge\bold{\underline{\underline{{Solution}}}} </p><p>Solution</p><p> </p><p>$

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..