Question

Find an ap whose 4 th term is 9 and sum of 6th term and 13th term is 40

Answer 1

Answer:

Step-by-step explanation:

Given,

a+3d=9......(I)

a+5d+a+12d=40

2a+17d=40.......(ii)

By applying elimination method to eqn I and ii

d=2

Now, put the value of d in eqn I

We have,

a+3d=9

a+3×2=9

a=3.

For an A.P.,

3,5,7,9,........

Answer 2

Step-by-step explanation:

4th term : 9 (a4= 9).

6th term + 13th term : 40 (a6+a13=40) ......(1)

a6 + a13 = 40 ( FROM 1)

a6 = a4 + 2d. .......... ......(2)

a13= a + 12d. ........ ........(3)

Applying (2) and (3) in (1)

(a4 + 2d ) + (a + 12d) = 40

a4+2d+a+12d=40

a4= a + 3d ( 4 )

applying ( 4 ) , we get

a +3d+ a + 12d + 2d = 40

2a + 17d = 40

$a \: + \: 3d \: = \: 9$

from above equations on elimination ,

we get ,

2a + 17d = 40

2a + 6d = 18

we get.. 11d = 22

d = 2....(A)

applying (A) in a+ 3d = 9

we get,

a+ 6 = 9

a = 3.

therefore , AP is

3 , 5 , 7 , 9 , 11....

hope this was helpful!