Question

if 4 and the last term of an ap are 11 and 89 respectively if there are 30 terms in the ap find the ap and its 23 terms

Answer 1

Hlo mate your solution

Answer 2

Answer:

The AP formed is 2 , 5 , 8 , 11 , . . . .

23rd term of AP is 68

Step-by-step explanation:

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 :

$a_{4}$ = 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 :

$a_{30}$ = 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 :

a + ( 23 - 1 ) d

= a + 22d

Substituting the value of a and d

= 2 + 22 ( 3 )

= 2 + 66

= 68

Hence,

The AP formed is 2 , 5 , 8 , 11 , . . . .

23rd term of AP is 68