Question

answer me sir madam plssss​

Answer 1

Given :-

• The sum of 4th and 8th term of AP is 24.
• The sum of 6th and 10th term of AP is 44.

To find :-

• First three terms of AP

Solution :-

Since the sum of 4th term (a4) and 8th term (a8) is 24,

=> a4 + a8 = 24

an can be written as a + (n-1)d

=> (a + 3d) + (a + 7d) = 24

=> 2a + 10d = 24

Divide both LHS and RHS with 2.

=> a + 5d = 24 -----Eqⁿ(1)

=> a6 = 24

This implies that 6th term of AP is 24.

Now we are given sum of 6th term (a6) and 10th term (a10) to be equal to 44.

=> a6 + a10 = 44

=> 24 + a10 = 44 (a6 = 24)

=> a10 = 44 - 24

=> a10 = 20

=> a + 9d = 20 -----Eqⁿ(2)

Now use equation (1) and (2) and eliminate a to get value of d.

Substracting Eqn(1) from (2)

Eqn(2) - Eqn(1)

=> a + 9d - (a + 5d) = 20 - 24

=> a + 9d - a - 5d = -4

=> 4d = -4

=> d = -1

Put this value of d in Eqn (2)

=> a + 9d = 20

=> a + 9(-1) = 20

=> a - 9 = 20

=> a = 29

Since we have obtained value of a = 29 and d = -1 , it's so simple to find first three terms of AP.

First term = a = 29

Second term = a+d => 29-1 => 28

Third term = a+2d => 29 - 2 => 27

So the required first three terms of AP are 29,28 and 27.