Question

in an arithmetic progression the 4th terms is 3 times the first term and the 7th term is one more than the twice the third term find the sequence

Answer 1

let the first term be a and common difference be d
according to the question,
T₄=3T₁
a+3d=3(a)
2a-3d=0...........1
again according to the question
T₇=2T₃+1
a+6d=2(a+2d)+1
a-2d=-1...........2
on solving 1 and 2
d=2 and a=3
the sequence is 3,5,7,9............

Answer 2

Solution :
Given Arithmetic progession
Let a be the first term then,
we know that,tn = a + (n-1) d


1) the 4th terms is 3 times the first term

t4 = 3a
a + 3d = 3a
3d = 2a

  thus  d= 2a/3

2)The 7th term is one more than the twice the third term

t7 = 2 ( t3 ) + 1
a +6d = 2 ( a + 2d ) + 1
substitute d= 2a/3
a + 6(2a/3) = 2 ( a + 4a/3 ) +1
a + 4a = 2 ( 7a/3 ) + 1
5a = 14a/3 + 1
5a - 14a/3 = 1
(15a - 14a )/3 = 1
a/3 =1
a = 3
hence d =2

the sequence is 3 , 5 , 7 , 9 , 11 , 13 , 15