Question

Find the 11th term of an arithmetic progression whose 4th term is 8 and the seventh term is 2 more than thrice of its third Tum

Answer 1

EXPLANATION.

• GIVEN

4th term of an Ap = 8

7th term is 2 more than thrice of it's third term.

To find 11th term.

According to the question,

Nth term of an Ap

=> An = a + ( n - 1 ) d

case = 1

=> 4th term = 8

=> a + 3d = 8 ....... (1)

case = 2

=> 7th term is 2 more than thrice of it's

third term.

=> a + 6d = 3 ( a + 2d ) + 2

=> a + 6d = 3a + 6d + 2

=> a - 3a = 2

=> -2a = 2

=> a = -1

put the value of a = -1 in equation (1)

we get,

=> a + 3d = 8

=> -1 + 3d = 8

=> 3d = 9

=> d = 3

Therefore,

First term = a = -1

common difference = d = 3

11th term of an Ap

=> a + 10d

=> -1 + 10(3)

=> -1 + 30

=> 29

Therefore,

11th term of an Ap = 29