Question

the sum of 4th and 8th term of an ap is 24 and the sum of the 6th and 10th term is 44. find the first three terms in ap

Answer 1

$T_4 + T_8 = 24 \\ \\ = a + 3d + a + 7d = 24 \\ \\ = 2a + 10d = 24 \\ \\ = 2(a + 5d) = 24 \\ \\ \\ a + 5d = \frac{24}{2} = 12 \\ \\ = T_6 = 12 \\ \\ \\ T_6 + T_{10} = 44 \\ \\ = 12 + T_{10} = 44 \\ \\ \\ T_{10} = 44 - 12 = 32$

$T_{10} - T_6 = 32 - 12 \\ \\ = (10 - 6)d = 20 \\ \\ = 4d = 20 \\ \\ \\ d = \frac{20}{4} = 5 \\ \\ \\ T_6 - T_1 = (6 - 1)d \\ \\ = 12 - T_1 = 5 \times 5 \\ \\ = 12 - T_1 = 25 \\ \\ \\ T_1 = 12 - 25 = -13 \\ \\ T_2 = -13 + 5 = -8 \\ \\ T_3 = -8 + 5 = -3$

∴ -13, -8 and -3 are the first three terms of the AP.

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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.