Question

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

Answer 1

Given:

${a}_{4} = a + 3d \\ {a}_{8} = a + 7d$

$a + 3d + a + 7d = 24 \\ \implies 2a + 10d = 24 \\ \implies a + 5d = 12 \bold{(Equation 1)}$

Now,

${a}_{6} = a + 5d \\ {a}_{10} = a + 9d$

$a + 5d + a + 9d = 44 \\ \implies 2a + 14d = 44 \\ \implies a + 7d = 22 \bold{(Equation 2)}$

Subtracting Equation 1 from Equation 2

$a + 7d - a - 5d = 22 - 12 \\ \implies 2d = 10 \\ \implies d = 5$

Putting value in Equation 2

a + 7d = 22

a + 7(5) = 22

a + 35 = 22

a = 22 - 35

a = -13

We know,

$a = {a}_{1}$

So,

${a}_{1} = -13 \\ {a}_{2} = a + d = -13 + 5 = -8 \\ {a}_{3} = a + 2d = -13 + 2(5) = -3$

Therefore, first three terms of AP are -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.