Question

the 4th and the last term of an ap are 11 & 89. IF there are 30terms in the AP find the AP and its 23 rd term

Answer 1

Dear student,

We know, The nth term in AP is given as : a_n = a + ( n - 1 ) d, where d is the common difference
According to question, 4th term of AP is 11
4 th term in AP can be written as : a₄ = a + ( 4 - 1 ) d = a + 3d
So, a + 3d = 11 [ Let this be Eqn 1 ]
Similarly,
Last term of AP is 89 , according to question, 30 th term in AP can be written as :
a₃₀ = a + ( 30 - 1 ) d = a + 29d
So,
a + 29d = 89 [ Let this be Eqn 2 ]
Subtracting Eqn 1 from Eqn 2 ,
a + 29d - ( a + 3d ) = 89 - 11
a + 29d - a - 3d = 78
26d = 78
d = 3
Substituting the value of d in Eqn 1 ,
a + 3 ( 3 ) = 11
a + 9 = 11
a = 11 - 9
a = 2
So, The first term of AP is 2 with difference between terms = 3
Therefore,
AP can be formed as : 2 , 5 , 8 , 11 , . . . .
23rd term of AP can be written as : a₂₃ = a + ( 23 - 1 ) d = a + 22d
Substituting the value of a and d
a₂₃ = 2 + 22 ( 3 ) = 2 + 66 = 68
Hence,
The AP formed is 2 , 5 , 8 , 11 , . . . .
23rd term of AP is 68.


Regards