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

S4=12
n/2(2a+n-1)D=12
12/2(2a+12-1)D=12
6(2a+11D)=12
2a+11d=2

A3+A4=14
a+2d+a+3d=14
2a+5d=14
substracting eq1 from 2
2a+5d=14.
2a+11d=2
(-)( - )=(-)
(-6d)=12)
d= (-2)


2a+11(-2)=2
2a-22=2
2a=2+22
2a=24
a=12

A2=a+d
=12+(-2)
=10
A3=a+2d
=12+2(-2) .
=12-4
=8