Question

find the first four consecutive terms in A.P whose sum is 12 and the sum of 3rd and 4th term is 14

Answer 1

Heya friend,


Here is your answer,

Let the four consecutive terms in A.P. be= a-3d, a-d, a+d, a+3d

Now,
According to the first condition,

a-3d + a-d + a+d + a+3d = 12
=> 4a = 12
=> a = 12/4
=> a = 3 ......( eq.1)

Now,
According to the second condition,

a+d+a+3d = 14
∴ 2a+4d = 14

=> 2(3) + 4d = 14 (from eq.1)
=> 6 + 4d = 14
=> 4d = 14 - 6
=> 4d = 8
=> d = 8/4
=> d = 2
=> a - 3d = 3-3(2) = 3 - 6 = -3

a-d = 3-2 = 1
a+d = 3+2 = 5
a+3d = 3+2(3) = 9

Therefore,
The four consecutive terms of A.P. are -3, 1, 5 and 9.


Hope it helps you.
Thank you.

Answer 2

Let the terms be a, a+d, a+2d, a+3d
Sum of the 4 terms = 12
⇒ a+a+d+a+2d+a+3d = 12
⇒ 4a + 6d = 12 → A
Sum of 4th term and 3rd term = 14
a+2d+a+3d = 14
2a+5d = 14 → B
Simultaneous equation A and B
2(4a+6d = 12)
4(2a+5d = 14)
⇒ 8a+12d = 24
    8a+20d = 56
 On subtracting B from A
  8a+12d = 24
- 8a-20d = -56
 ⇒ - 8d =  -32
⇒ d = 4
8a + 12(4) = 24
8a + 48 = 24
8a = -24
a = -3
The sequence of terms are -3, 1, 5, 9