Question

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


Please don't give incorrect answers i need help​

Answer 1

Answer:

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Given ,

Sum of 4th and 8th term is 24 .

$a + 3d + a + 7d = 24$

$2a + 10d = 24$

$a + 5d3 = 12.......(1)$

Then ,

Sum of 6th and 10th term is 44.

$a + 5d + a + 9d = 44$

$12 + a + 9 d = 44......from \: (1)$

$a + 9d = 32$

$a + 5d + 4d = 32$

$12 + 4d = 32$

$d = 5$

$a + 25 = 12$

$a = - 13$

AP = -13 , -8 , -3 , 2 , 7 , 12 ............

Answer 2

Solution:

4th term =a+3d

8th term =a+7d

sum of the 4th term and 8th term=a+3d+a+7d=24

⇒2a+10d=24

Take 2 common from the equation a+5d=12-----(1)

sum of the 6th term and 8th term=44

⇒a+5d+a+9d=44

2a+14d=44

Take 2 common from equation

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

By elimination Method:-

a+5d=12

a+7d=22

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2d=10

d=10/2

d=5

Substitute d=5 in equation (1)

a+5d=12

a+5(5)=12

a=12-25

a= -13

⇒The 1st 3 terms are :- -13,-13,+5,-13,+10

⇒-13,-8,-3.

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