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:

Answer 2

Answer:

The AP is defined as:

First term = a = 3

Common difference = d = 2

Step-by-step explanation:

For an AP, let

a = first term

d = common difference

Then,

$a_{n}$ = nth term = a + (n - 1)d

Given:

$a_{4}$ = 9

=> a + (4 - 1)d = 9

=> a + 3d = 9 ....(i)

Also,

$a_{6} + a_{13} = 40$

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

=> a + 5d + a + 12d = 40

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

Multiplying (i) by 2, we get:

2a + 6d = 18 .......(iii)

Subtracting (iii) from (ii), we get:

11d = 22

=> d = 2

Substituting for d in (i), we get:

a + 3(2) = 9

=> a + 6 = 9

=> a = 3

The AP is defined as:

First term = a = 3

Common difference = d = 2

Verification:

The AP will be:

3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, ...

We can see that:

fourth term is 9 √

sum of sixth term and thirteenth term is 40 √