Question

the sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first 3 terms of the AP ​

Answer 1

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Given:

a4+a8=24

Now,

a+3d+a+7d=24

2a+10d=24

a+5d=12------(1)

Also,

a6+a10=44

a+5d+a+9d=44

2a+14d=44

a+7d=22-----(2)

Subtracting (2) from (1),

-2d=-10

d=5

Substituting d=5 in (1),

a+5(5)=12

a+25=12

a=12-25

a=-13

Therefore, the first three terms of the Arithmetic progression are:

a=-13

a+d=-13+5==-8

a+2d=-13+2(5)=-13+10=-3

Answer: -13,-8,-3....

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.