Question

sum of the first 15 terms of an ap is 405 and sum of first 21 terms is 756. find the 9 th term and common difference​

Answer 1

Given :

• sum of the first 15 terms of an ap is 405 and sum of first 21 terms is 756

To find :

• find the 9 th term and common difference

Solution :

let the first term be a and the commen difference be d

now it is given that,

s_15 = 15/2[2a + (15 - 1)d] = 405

405 = 15/2 [2a + 14d]

a + 7d = 27 .......... ( I )

and,

s_21 = 21/2[2a + (21 - 1)d] = 756

756 = 21/2[2a + 20d]

a + 10d = 36 .......( ii )

equation (I) subtracting from equation (ii) we get,

a + 10d = 36

- a + 7d = 27

----------------------

⠀⠀⠀3d = 9

d = 9/3

d = 3

from equation (I)

a + 7d = 27

a = 27 - 7 x 3

a = 27 - 21

a = 6

t_n = a + (n - 1) d

t_9 = 6 + (9 - 1) 3

t_9 = 6 + 24

t_9 = 30

t_9 = 30

∴ The 9 th term is 30 and common difference 3

Answer 2

Given :

sum of the first 15 terms of an ap is 405 and sum of first 21 terms is 756

To find :

find the 9 th term and common difference

Solution :

let the first term be a and the commen difference be d

now it is given that,

s_15 = 15/2[2a + (15 - 1)d] = 405

405 = 15/2 [2a + 14d]

a + 7d = 27 .......... ( I )

and,

s_21 = 21/2[2a + (21 - 1)d] = 756

756 = 21/2[2a + 20d]

a + 10d = 36 .......( ii )

equation (I) subtracting from equation (ii) we get,

a + 10d = 36

- a + 7d = 27

----------------------

⠀⠀⠀3d = 9

d = 9/3

d = 3

from equation (I)

a + 7d = 27

a = 27 - 7 x 3

a = 27 - 21

a = 6

s_n = n/2 [2a + (n - 1) d]

s_9 = 9/2 [2 x 6 + (9 - 1) 3]

s_9 = 9/2 (12 + 24)

s_9 = 9 x 18

s_9 = 162

∴ The 9 th term is 162 and common difference 3