Question

in an Arithmetic progression the 4th term is 3 times the first term and the 7th term is one more than the twice the third term find the sequenve

Answer 1

t4=3*t1
t7=1+t3
so 
a+3d=3a.......................1
a+6d=1+a+2d.....................2
2a=3d
4d=1
so d=1/4
2a=3/4
a=3/8
s0
3/8,5/8,7/8,9/8,....................

Answer 2

let the first term be a and common difference be d
according to the question
t₄=3*t₁
a+3d=3a
2a-3d=0.............1

t₇=2t₃+1
a+6d=2(a+2d)+1
a+6d=2a+4d+1
a-2d=1..............2

on solving 1 and 2
we get 
a=3 and d=2
therefore the sequence is 3,5,7,9.................