Math, asked by Mister360, 2 months ago

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

Answers

Answered by thebrain2312
2

Answer:

Here,

a

4

=a+3d

a

8

=a+7d

Therefore,

a+3d+a+7d=24

2a+10d=24

a+5d=12 …… (1)

Again,

a

6

=a+5d

a

10

=a+9d

Therefore,

a+5d+a+9d=44

2a+14d=44

a+7d=22 ……. (2)

Solving equations (1) and (2), we get

d=5 and a=−13

Therefore,

a

1

=a=−13

a

2

=a+d=−13+5=−8

a

3

=a+2d=−13+10=−3

Step-by-step explanation:

a

4

=a+3d

a

8

=a+7d

Therefore,

a+3d+a+7d=24

2a+10d=24

a+5d=12 …… (1)

Again,

a

6

=a+5d

a

10

=a+9d

Therefore,

a+5d+a+9d=44

2a+14d=44

a+7d=22 ……. (2)

Solving equations (1) and (2), we get

d=5 and a=−13

Therefore,

a

1

=a=−13

a

2

=a+d=−13+5=−8

a

3

=a+2d=−13+10=−3

Answered by Anonymous
20

Answer :-

We know that :-

Nth term of AP -

aₙ = a + ( n - 1 ) d

So,

4th term of AP = a + 3d

8th term of AP = a + 7d

Sum of 8th and 4th term = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12

Similarly,

6th term of AP = a + 5d

10th term of AP = a + 9d

Sum of 6th and 10th term = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22

Now, we have 2 equations -

⇒ a + 5d = 12 ( i )

⇒ a + 7d = 22 ( ii )

Subtracting equation i from ii

⇒ a + 7d - a - 5d = 22 - 12

⇒ 2d = 10

⇒ d = 10/2

⇒ d = 5

Substituting the value in equation i

⇒ a + 5d = 12

⇒ a + 5 × 5 = 12

⇒ a = 12 - 25

⇒ a = -13

Now we have,

  • d = 5
  • a = - 13

First term = -13

Second term = a + d

= -13 + 5

= -8

Second term = -8

Third term = a + 2d

= -13 + 2 × 5

= -13 + 10

= -3

Third term = -3

First three terms of AP are - 13 , - 8 and - 3.

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