Question

Find the AP whose 4th term is 9 and the sum of 6 th term and 13th term is 40

Answer 1

Answer:

The AP required here is 3 , 5 , 7 , 9 , 11 .....

Step-by-step explanation:

Let a be the first term of the AP and d be the common difference between the terms.

It is given that the 4th term of the AP is 9.

So,

= >  a + ( 4 - 1 ) d = 9

= >  a + 3d = 9

= >  a = 9 - 3d       ...( i )

Also, given that the sum of 6th and 13th term of the AP is 40.

So,

= >  6th term + 13th term = 40

= >  a + ( 6 - 1 )d + a + ( 13 - 1 )d = 40

= >  a + 5d + a + 12d = 40

= >  2a + 17d = 40

Substituting the value of a from ( i ),

= >  2( 9 - 3d ) + 17d = 40

= >  18 - 6d + 17d = 40

= >  18 + 11d = 40

= >  11d = 40 - 18

= >  11d = 22

= >  d = 22 / 11

= >  d = 2

Now, substitute the value of  d in ( i ),

= >  a = 9 - 3d

= >  a = 9 - 3( 2 )

= >  a = 9 - 6

= >  a = 3

So,

Required AP : a , a + d , a + 2d..... = 3 , 3 + 2 , 3 + 2( 2 ) ..... = 3 , 5 , 7 , 9.....