Question

18. The sum of the 4th and 8th terms ofan AP is 24 and the sum of the 6th and 10th terms is
44. Find the first three terms of the AP
19​

Answer 1

Answer:

-24

Step-by-step explanation:

an=a+(n-1)d

a4+a8=24

a+3d+a+7d=24

2a+10d=24

1)a+5d=12

a6+a10=44

a+5d+a+9d=44

2)a+7d=22

subtract the two equations

we get d=5,a=-13

sum of first three terms of an AP=a1+a2+a3

=a+a+d+a+2d

=-24

Answer 2

Answer:

The first three terms of an AP are -13,-8,-3.

Step-by-step explanation:

Given :-

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

To find :-

• The first three teams of the AP.

Solution :-

Consider,

The first three terms of an AP,

a , a+d , a+2d

Here,

• a = first term
• d = common difference

Formula used :-

${\boxed{\sf{T_n=a+(n-1)d}}}$

Then,

★$\sf{T_4=a+(4-1)d}$

$\to\sf{T_4=a+3d}$

And

★$\sf{T_8=a+(8-1)d}$

$\to\sf{T_8=a+7d}$

According to the 1st condition,

$\sf{T_4+T_8=24}$

$\to\sf{a+3d+a+7d=24}$

$\to\sf{2a+10d=24............(i)}$

Similarly,

★$\sf{T_6=a+(6-1)d}$

$\to\sf{T_6=a+5d}$

And

★$\sf{T_{10}=a+(10-1)d}$

$\to\sf{T_{10}=a+9d}$

According to the 2nd condition,

$\sf{T_6+T_{10}=44}$

$\to\sf{a+5d+a+9d=44}$

$\to\sf{2a+14d=44............(ii)}$

Now subtract eq(ii) from eq(i).

2a+10d-(2a+14d)=24-44

→ 2a+10d-2a-14d=-20

→ -4d=-20

→ d=5

Now put d=5 in eq(i) for getting the value of a.

2a + 10×5 = 24

→ 2a + 50 = 24

→ 2a = -26

→ a = -13

Therefore,

The first three terms of an AP ,

• a = -13
• a+d = -13+5 = -8
• a+2d = -13+2×5 = -3