Question
The fourth term of an A.P. is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is
A.
$$2$$
B.
$$3$$
C.
$$\frac{3}{2}$$
D.
$$- 1$$
Answer :
$$2$$
Solution :
Let the progression be $$a, a + d, a + 2d$$
Then $${x_4} = 3{x_1}$$
$$\eqalign{
& \Rightarrow a + 3d = 3a \cr
& \Rightarrow 3d = 2a\,\,\,\,\,.....\left( {\text{i}} \right) \cr
& {\text{Again, }}{x_7} = 2{x_3} + 1 \cr
& \Rightarrow a + 6d = 2\left( {a + 2d} \right) + 1 \cr
& \Rightarrow 2d = a + 1\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Solving (i) and (ii) we get
$$a = 3, d = 2$$