Question

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
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Answer 1

Answer:

a = -13 ; d = 5

Step-by-step-explanation:

Given :

4th term + 8th term = 24 ;

6th term + 10th term = 44 ;

》4th term ( t4 ) = a+3d ; 8th term ( t8 ) = a+ 7d

=> a+3d + a+7d = 24 [ According to the question ]

=> 2a+10d = 24

=> a+5d = 12 ( On dividing by 2 ) --> eq ①

》6th term ( t6 ) = a+5d ; 10th term ( t10 ) = a+ 9d

=> a+5d + a+9d = 44 [ According to the question ]

=> 2a+14d = 44

=> a+7d = 22 ( On dividing by 2 ) --> eq ②

》Subtract eq ② from eq ①

a+7d = 22

a+5d = 12

________

2d = 10

________

=>${d = {\frac{10}{2}}}$

=> ${\boxed{d = 5}}$

》Substitute d = 5 in eq ①

=> a + 5 (5) = 12

=> a = 12 - 25

=>${\boxed{a = -13}}$

》First 3 terms =

a, a+d, a+2d ....

=> -13, -13+5, -13+2 (5)

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Answer 2

Step-by-step explanation:

a+3d+a+7d=24

2a+10d=24

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

sum of 6 th and 10 th term is 44.

a+5d+a+9d=44

12+a+9d=44......from (i)

a+9d=32

a+5d+4d=32

12+4d=32

d=5

a+25=12

a=−13

AP=−13,−8,−3,2,7,12....