Question

Answer the following question :-
1) The sum of the 4th and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44.
Find the first three terms of the AP.​

Answer 1

Answer:-

-8 , -13 , -18

Given :-

$a_4 + a_8 = 24$

$a_6 + a_{10} = 44$

To find:-

The three term in A.P

Solution:-

Let a be the first term and d be the common Difference.

A/Q

$a_4 + a_8 = 24$

→$a + 3d + a + 7d = 24$

→$2a + 10d = 24$

Divide both side by 2 .

→$\dfrac{2a}{2}+ \dfrac{10d}{2}= \dfrac{24}{2}$

→$a + 5d = 12 -----eq.1$

Now,

$a_6 + a_{10} = 44$

→$a + 5d + a + 9d = 44$

→$2a +14d = 44$

Divide both side by 2.

→$\dfrac{2a}{2}+\dfrac{14d}{2}=\dfrac{44}{2}$

→$a + 7d = 22 ------eq.2$

Subtract equation1 and 2 we get,

→$a -a + 5d -7d = 12 -22$

→$-2d = -10$

→$d = \dfrac{-10}{-2}$

→$d = 5$

put the value in equation 1.

a + 5d = 12

a + 5 × 5 = 12

a + 25 = 12

a = 12 - 25

a = -13

Now , three term in AP is given by :-

→a = -13

→a + d = -13 + 5 = -8

→a - d = -13 - 5 = -18

Answer 2

Here, t4+t8=24. or,a+3d+a+7d=24.

or,2a+10d=24.

or,2a=24-10d.

or 2(a)=2(12-5d) So,a=12-5d-----equation 1st. And also,

t6+t10=44.

or,a+5d+a+10d=44.

or,2a+15d=44.

or,2(12-5d)+15d=44.(From equation 1st)

or,24-10d+15d=44.

or,5d=44-24.

or,5d=20

Therefore,d=4.

Again,From equation 1st. a=12-5d=12-5×4=12-20=-8

So,d=4 and a=-8

Then,The first three terms of ap are

t1=a=-8

t2=a+d=-8+4=-4

t3=a+2d=-8+2×4=0

Step-by-step explanation:

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