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

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) (3/2) (D) -1

Tardigrade Admin · Accepted Answer

Let the progression be a, a+d, a+2 d, Then x4=3 x1 ⇒ a+3 d=3 a ⇒ 3 d=2 a ldots (i) Again x7=2 x3+1 ⇒ a+6 d=2(a+2 d)+1 ⇒ 2 d=a+1 ldots (ii) Solving (i) and (ii), we get a =3, d =2

Q. 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

2

3

$\frac{3}{2}$

-1

Let the progression be $a, a + d, a + 2 d$ ,
Then $x_{4} = 3 x_{1}$
$\Rightarrow a + 3 d = 3 a$
$\Rightarrow 3 d = 2 a \dots$ (i)
Again $x_{7} = 2 x_{3} + 1$
$\Rightarrow a + 6 d = 2 (a + 2 d) + 1$
$\Rightarrow 2 d = a + 1 \dots$ (ii)
Solving (i) and (ii), we get
$a = 3, d = 2$

Q. 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

2

3

23​

-1

Let the progression be a,a+d,a+2d, Then x4​=3x1​ ⇒a+3d=3a ⇒3d=2a… (i) Again x7​=2x3​+1 ⇒a+6d=2(a+2d)+1 ⇒2d=a+1… (ii) Solving (i) and (ii), we get a=3,d=2

$\frac{3}{2}$

Let the progression be $a, a + d, a + 2 d$ ,
Then $x_{4} = 3 x_{1}$
$\Rightarrow a + 3 d = 3 a$
$\Rightarrow 3 d = 2 a \dots$ (i)
Again $x_{7} = 2 x_{3} + 1$
$\Rightarrow a + 6 d = 2 (a + 2 d) + 1$
$\Rightarrow 2 d = a + 1 \dots$ (ii)
Solving (i) and (ii), we get
$a = 3, d = 2$