Question

the sum of the forth and fifth term of a AP is 24 and the sum of sixth and tenth is 44 find the first three terms of AP​

Answer 1

Answer:

Here,

a

4

=a+3d

a

8

=a+7dTherefore,

a+3d+a+7d=24

2a+10d=24

a+5d=12 …… (1)Again,

a

6

=a+5d

a

10

=a+9dTherefore,

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

Hence, this is the required result.

Answer 2

ANSWER:

1st = 2

2nd = 68/40

EXPLANATION:

Let first term of the AP be x

And common difference be d

therefore,

4th term = a + 3d

5th term = a + 4d

6th term = a + 5d

10th term = a +9d

NOW, 4th + 5th = 24

so, a + 3d + a + 4d = 24

2a + 7d = 24. (i)

so, 4a +14d = 48 (ii). (multiplied the above equation by 2)

AND, 6th + 10th = 44

so, a + 5d + a + 9d = 44

2a + 14d = 44. (iii)

subtracting (iii) from (i)

4a + 14d - 2a - 14d = 48 - 44

2a = 4

a = 2

now put a's value in (ii)

4a + 14d = 48

4(2) + 14d = 48

14d = 40

d = 40/14

now, 1st term = a = 2

and, 2nd term = a + d = 2 + 40/14 = 68/40