Question

The sum of the 4th and 8th terms of an ap is 24 and the sum of the 6th and 10th terms is 44 find the 1st twelve terms​

Answer 1

Answer:

-13, - 8, - 3, 2.....

Explanation:

Let the terms of an A.P be a, (a + d), (a + 2d), (a + 3d), (a + 4d),.......

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

➾ a + 3d + a + 7d = 24

➾ 2a + 10d = 24

➾ 2(a + 5d) = 24

➾ a + 5d =24/2

➾ a + 5d = 12 ....... (i)

Also,

Sum of 6th and 10th term is 44.

➾ a + 5d + a + 9d = 44

➾ 2a + 14d = 44

➾ 2(a + 7d) = 44

➾ a + 7d =44/2

➾ a + 7d = 22 ..... (ii)

Subtracting (i) and (ii)

➾ (a - a) + (7d - 5d) = (22 - 12)

➾ 2d = 10

➾ d = 10/2

➾ d = 5

Substituting the value of d in (i) -

➾ a + 5d = 12

➾ a + 5 * 5 = 12

➾ a = 12 - 25

➾ a = - 13

So, answer is - 13, - 8, - 3, 2.....

Answer 2

Answer:

$\large{\underline{\boxed{\sf - 13, - 8, - 3, 2, 7, 12, 17, 22, 27, 32, 37, 42..}}}$

Explanation:

Let the terms of an A.P be a, (a + d), (a + 2d), (a + 3d), (a + 4d),.......(a + nd) respectively.

Given that ;

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

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ 2(a + 5d) = 24

⇒ a + 5d = $\sf \dfrac{24}{2}$

⇒ a + 5d = 12 ....... (i)

Also,

Sum of 6th and 10th term is 44.

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44

⇒ 2(a + 7d) = 44

⇒ a + 7d = $\sf \dfrac{44}{2}$

⇒ a + 7d = 22 ..... (ii)

_______________________

Now, Subtracting equation (i) from (ii) -

⇒ (a - a) + (7d - 5d) = (22 - 12)

⇒ 2d = 10

⇒ d = $\sf \dfrac{10}{2}$

⇒ d = 5

Substituting the value of d in (i) -

⇒ a + 5d = 12

⇒ a + 5 * 5 = 12

⇒ a = 12 - 25

⇒ a = - 13

_______________________

So, the terms of A.P are ;

• a₁ = - 13
• a₂ = - 13 + 5= - 8
• a₃ = - 8 + 5 = - 3
• a₄ = - 3 + 5 = 2
• a₅ = 2 + 5 = 7
• a₆ = 7 + 5 = 12
• a₇ = 12 + 5 = 17
• a₈ = 17 + 5 = 22
• a₉ = 22 + 5 = 27
• a₁₀ = 27 + 5 = 32
• a₁₁ = 32 + 5 = 37
• a₁₂ = 37 + 5 = 42