Question

If sum of 4th nd 8th term of ap is 24 and sum of 6th nd 8th term is 44 find Ap​

Answer 1

Step-by-step explanation:

Given:

$t_4+t_8=24 \\ \therefore \: a + 3d + a + 7d = 24 \\ \implies \: 2a + 10d = 24 \\ \implies \: 2(a + 5d) = 24 \\ \implies \: a + 5d =12.....(1) \\ next \\ t_6+t_8=44.... (Given) \\ \therefore \: a + 5d + a + 7d = 44 \\ \implies \: 2a + 12d = 44 \\ \implies \: 2(a + 6d) = 44 \\ \implies \: a + 6d =22.....(2) \\ subtrcting \: (1) \: from \: (2) \\ a + 6d - (a + 5d)=22 - 12 \\ a + 6d - a - 5d=10 \\ \implies \: \huge \fbox{d = 10} \\ from \: equation \: (1) \\ a + 5 \times 10 = 12 \\ a + 50 = 12 \\ a = 12 - 50 \\ \implies \huge \fbox{a = - 38} \\ thus \\ t_1 = a = - 38 \\ t_2 = t_ 1 + d = - 38 + 10 = - 28 \\ t_3 = t_ 2 + d = - 28 + 10 = - 18 \\ t_4 = t_ 3 + d = - 18 + 10 = - 8 \\ t_5 = t_ 4 + d = - 8 + 10 = 2 \\ t_6 = t_ 5 + d = 2 + 10 = 12 \\ Hence \: the \: required \: AP \: is : \\ - 38, \: - 28, \: - 18, \: - 8, \: 2, \: 12, \: .............$

Answer 2

Answer:

- 13 , - 8 , - 3 .

Step-by-step explanation:

Let the first term a and common difference be d.

We know :

t_n = a + ( n - 1 ) d

t_4 = a + 3 d

t_8 = a + 7 d

We have given :

t_4 + t_8 = 24

2 a + 10 d = 24

a + 5 d = 12

a = 12 - 5 d ....( i )

t_6 = a + 5 d

t_10 = a + 9 d

: t_6 + t_10 = 44

2 a + 14 d = 44

a + 7 d = 22

a = 22 - 7 d ... ( ii )

From ( i ) and  ( ii )

12 - 5 d = 22 - 7 d

7 d - 5 d = 22 - 12

2 d = 10

d = 5

We have :

a = 12 - 5 d

a = 12 - 25

a = - 13

Now required answer as :

- 13 , - 8 , - 3 .

Finally we get answer.