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 first 3 terms​

Answer 1

3 Terms = – 13 , – 8 , – 3

Step-by-step explanation:

Given:

• Sum of 4th and 8th terms of AP is 24.
• Sum of 6th and 10th terms is 44.

To Find:

• What are first three terms of AP ?

Solution: Let the first term of AP be a and common difference be d.

As we know that an AP series is given by

aⁿ = a + (n – 1)d

∴ 4th term of AP = a + (4 – 1)d

∴ 8th term of AP = a + (8 – 1)d

A/q

• 4th + 8th term = 24

➟ a + 3d + a + 7d = 24

➟ 2a + 10d = 24

➟ 2(a + 5d) = 24

➟ a + 5d = 24/2

➟ a = (12 – 5d).........1

Now, Sum of 6th and 10th terms is 44.

∴ 6th term = a + (6 – 1)d

∴ 10th term = a + (10 – 1)d

Sum is 44. Therefore,

➟ a + 5d + a + 9d = 44

➟ 2a + 14d = 44

➟ 2(a + 7d) = 44

➟ a + 7d = 44/2

➟ 12 – 5d + 7d = 22 {from equation 1}

➟ 2d = 22 – 12

➟ d = 10/2 = 5

We got the value of d = 5. Now put the value of d in equation 1

• a = 12 – 5(5) = 12 – 25 = – 13

Now, we got a = – 13 & d = 5. So required terms will be

• a¹ = a = – 13

• a² = a + d = – 13 + 5 = – 8

• a³ = a + 2d = – 13 + 2(5) = – 3

Answer 2

Solution :

$\underline{\bf{Given\::}}$

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

$\underline{\bf{Explanation\::}}$

As we know that formula of an A.P;

$\boxed{\bf{a_n=a+(n-1)d}}$

• a is the first term.
• d is the common difference.
• n is the term of the number.

A/q

$\longrightarrow\sf{a_4 + a_8 = 24}\\\\\longrightarrow\sf{a+(4-1)d + a+(8-1)d=24}\\\\\longrightarrow\sf{a+3d + a+7d =24}\\\\\longrightarrow\sf{2a + 10d =24}\\\\\longrightarrow\sf{2(a+5d ) = 24}\\\\\longrightarrow\sf{a+5d=\cancel{24/2}}\\\\\longrightarrow\sf{a+5d=12}\\\\\longrightarrow\sf{a=12-5d......................(1)}$

&

$\longrightarrow\sf{a_6 + a_{10} = 44}\\\\\longrightarrow\sf{a+(6-1)d + a+(10-1)d=44}\\\\\longrightarrow\sf{a+5d + a+9d =44}\\\\\longrightarrow\sf{2a + 14d =44}\\\\\longrightarrow\sf{2(a+7d ) = 44}\\\\\longrightarrow\sf{a+7d=\cancel{44/2}}\\\\\longrightarrow\sf{a+7d=22}\\\\\longrightarrow\sf{12-5d + 7d = 22\:\:[from(1)]}\\\\\longrightarrow\sf{12+2d = 22}\\\\\longrightarrow\sf{2d=22-12}\\\\\longrightarrow\sf{2d=10}\\\\\longrightarrow\sf{d=\cancel{10/2}}\\\\\longrightarrow\bf{d=5}$

∴ Putting the value of d in equation (1),we get;

$\longrightarrow\sf{a=12-5(5)}\\\\\longrightarrow\sf{a=12-25}\\\\\longrightarrow\bf{a=-13}$

Now;

A R I T H M E T I C  P R O G R E S S I O N  :

• a = -13
• a + d = -13 + 5 = -8
• a + 2d = -13 + 2 × 5 = -13 + 10 = -3

Thus;

The first 3 terms of an A.P. will be -13 , -8 & -3 .