Question

The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third time by 1 . Find the first term and the common difference.​

Answer 1

Step-by-step explanation:

Given:-

• The 4th term of an A.P. is three times the 1st term.

• The 7th term exceeds twice the 3rd term by 1.

To Find:-

• The first term.

• The Common difference.

Solution:-

Let " a " be the first term and " d " be the common difference.

Case 1:-

The 4th term is three times the 1st term.

$\sf\implies a_{4} = 3a$

$\sf\implies a + 3d = 3a$

$\sf\implies a - 3a + 3d = 0$

$\sf\implies -2a + 3d = 0.......(i)$

Case 2:-

The 7th term exceeds twice the 3rd term by 1.

$\sf\implies a_{7} = 2(a_{3} ) + 1$

$\sf\implies a + 6d = 2(a+ 2d ) + 1$

$\sf\implies a + 6d = 2a+ 4d + 1$

$\sf\implies a - 2a + 6d - 4d = 1$

$\sf\implies -a + 2d = 1.....(ii)$

Multiplying equation (ii) with 2

$\sf\implies -2a + 4d = 2.....(iii)$

Equation (iii) - (i)

$\sf\implies -2a + 4d - (-2a + 3d) = 2$

$\sf\implies -2a + 4d + 2a - 3d = 2$

$\sf\implies d = 2$

Substitute d = 2 in equation (ii)

$\sf\implies -a + 2d = 1.....(ii)$

$\sf\implies -a + 2(2) = 1$

$\sf\implies -a = 1 - 4$

$\sf\implies -a = -3$

$\sf\implies a = 3$

Therefore:-

$\large \underline{\boxed{\textsf{\textbf{The \: first \: term = 3}}}}$

$\large \underline{\boxed{\textsf{\textbf{Common \: difference = 2}}}}$

Answer 2

The nth term of an A.P is given by

a$\small{n}$=a$\small{1}$+(n-1)d

where d is the common difference

and a$\small{1}$is the first term of A.P

hence 4th term of an A.P is

a$\small{4}$=a$\small{1}$+(4-1)d

⟹a$\small{4}$=a$\small{1}$+3d.....eq(1)

given that 4th term of A.P is three times the first .

⟹a$\small{4}$=3a$\small{1}$

put value of aa$\small{4}$from eq(1)

⟹a$\small{1}$+3d=3a$\small{1}$

⟹3a$\small{1}$-a$\small{1}$=3d

⟹2a$\small{1}$=3d

⟹a$\small{1}$=$\dfrac{3d}{2}$....eq(2)

3rd term of A.P is given by

a$\small{3}$=a$\small{1}$+(3-1)d

a$\small{3}$=a$\small{1}$+2d

7th term of A.P is given by

a$\small{7}$=a$\small{1}$+6d

given that 7th term exceeds twice the third by 1

⟹a$\small{7}$=2a$\small{3}$+1

put values ofa$\small{7}$and a$\small{3}$

a$\small{1}$+6d=2(a$\small{1}$+2d)+1

⟹2a$\small{1}$-a$\small{1}$=6d-4d-1

⟹a$\small{1}$=2d-1....eq(3)

put value of a$\small{1}$- from eq(2) to eq(3)

$\dfrac{3d}{2}$=2d−1

⟹3d=4d−2

⟹4d−3d=2

⟹d=2...…..eq(4) common difference of A.P

put value of d from eq(4) to eq(2) we get

⟹a$\small{1}$=$\dfrac{3×2}{2}$

⟹a$\small{1}$=3...... first term of A.P