Question

for first four consecutive terms in AP whose sum is 12 and the sum of 3rd and 4th term is 14

Answer 1

let the terms be a-d, a,a+d,a+2d
sum=4a+2d=12
2a+d=6......(1)
sum of 3rd and 4th =(a+d)+(a+2d)=14
2a+3d=14............(2)
subtract (1)-(2)
-2d=-8
d=4
from (1)
a=1
terms are
(1-4),1,(1+4),(1+2(4))
-3, 1, 5, 9

Answer 2

Answer:

The terms are: -3, 1, 5, 9

Step-by-step explanation:  

Given that for first four consecutive terms in AP whose sum is 12 and the sum of 3rd and 4th term is 14.

we have to find the terms

Let the terms be a-d, a,a+d,a+2d

Sum=4a+2d=12

2a+d=6     →   (1)

Sum of 3rd and 4th =(a+d)+(a+2d)=14

2a+3d=14   →  (2)

Solving (1) and (2)

(2)-(1) ⇒ 2d=8

d=4

from (1), we get

2a+4=6 ⇒ a=1

Terms are:

(1-4), 1, (1+4), (1+2(4))

-3, 1, 5, 9