Question

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

Answer 1

Answer:

- 13 , - 8 and - 3

Step-by-step explanation:

nth term in APs is defined by a + ( n - 1 )d, where symbols are their usual meaning.  

    Let first term of this AP be a, and common difference be d.

Therefore,

      Sum of 4th and 8th term = 24

⇒ [ a + 3d ] + [ a + 7d ] = 24

⇒ 2a + 10d = 24

⇒ 2( a + 5d ) = 24

 ⇒ a + 5d = 12

 ⇒ a = 12 - 5d

       Sum of 6th and 10th term is 44

⇒ [ a + 5d ] + [ a + 9d ] = 44

⇒ 2a + 14d = 44

⇒ 2( a + 7d ) = 44

⇒ a + 7d = 22

        a = 12 - 5d

⇒ 12 - 5d + 7d = 22

⇒ 12 + 2d = 22

⇒ 2d = 22 - 12 = 10

⇒ d = 10/2 = 5

        Hence,

     a = 12 - 5d = 12 - (5)

        = 12 - 25

        = - 13

 Hence,

first term = - 13

2nd = a + d = - 13 + 5 = - 8

3rd = a + 2d = - 13 + 2(5) = - 3

Answer 2

☣ GIVEN ☣

↬ Sum of the 4th and 8th terms of an AP is 24.

↬ Sum of the 6th and 10th terms of the AP is 44.

☣ TO FIND ☣

» The first three terms of the AP.

☣ SOLUTION ☣

According to the question,

$\sf{a_4+a_8=24}$

$\sf{\longmapsto a+3d+a+7d=24}\\\\\sf{\longmapsto 2a+10d=24}\\\\\sf{\longmapsto a+5d=12}\:\:\:\:\:\:\:\: \cdots \sf{(i)}$

$\rule{130}2$

$\sf{a_6+a_{10}=44}$

$\sf{\longmapsto a+5d+a+9d=44}\\\\\sf{\longmapsto 2a+14d=44}\\\\\sf{\longmapsto a+7d=22}\:\:\:\:\:\:\:\: \cdots \sf{(ii)}$

$\rule{130}2$

Subtracting eq.(i) from eq.(ii) :-

$\sf{a+7d=22}\\-\\ \sf{a+5d=12}\\\rule{50}1\\\sf{2d=10}$

$\sf{\longmapsto d=\dfrac{10}{2} }\\\\\boxed{\sf{\longmapsto d=5}}$

Putting the value of d in eq.(i) :-

$\sf{a+5d=12}\\\\\sf{\longmapsto a+5(5)=12}\\\\\sf{\longmapsto a=12-25}\\\\\boxed{\sf{\longmapsto a=-13}}$

$\rule{130}2$

Now, the required AP is as follows :-

a₁ = a = -13

a₂ = (a + d) = -13 + 5 = -8

a₃ = (a + 2d) = -13 + 2(5) = -13 + 10 = -3

Therefore, the AP is -13, -8, -3, .....