Question

dentermine the Ap where 3rd term is 5 and 7th term is 9​

Answer 1

To Find :

• we need to find the AP

Solution :

• 3rd term of AP = 5
• 7th term of AP = 9

As we know that,

• an = a + (n - 1)d

Where,

• an = nth term
• a = first term
• n = number of terms
• d = common difference.

a3 = a + (3 - 1)d

• a3 = a + 2d .....1)

a7 = a + (7 - 1)d

• a7 = a + 6d ....2)

From equation (1) and (2)

we get,

⠀⠀⠀⠀a + 2d = 5

⠀⠀⠀⠀a + 6d = 9

⠀⠀⠀⠀--⠀⠀--⠀⠀--⠀⠀

⠀⠀⠀⠀⠀ - 4d = - 4

⠀⠀⠀⠀⠀⠀ d = - 4/-4

⠀⠀⠀⠀⠀⠀⠀d = 1

Substituting value d in equation (1)

a + 2d = 5

a + 2 × 1 = 5

a = 5 - 2

• a = 3

So,

• AP is :-

a = 3

a + d = 3 + 1 = 4

a + 2d = 3 + 2 × 1 = 5

a + 3d = 3 + 3 × 1 = 6

• Hence the AP is 3 , 4 ,5 ,6 ....

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Answer 2

Solution:-

Given

$\circ \rm \: T_3 = 5$

$\circ \rm \: T_7= 9$

Formula

$\boxed{ \to \: \rm T_n = a + (n - 1)d}$

We get

$\circ \rm \: Take \: T_3 = 5$

$\rm \: 5 = a + (3 - 1)d$

$\rm \: a + 2d = 5 \: \: \: \: \: \: ......(i)eq$

$\circ \rm \: Take \: T_7 = 9$

$\rm \: 9 = a + (7 - 1)d$

$\rm \: a + 6d = 9 \: \: \: \: \: \: \: \: .......(ii)eq$

Using elimination method

Subtracting (ii) with (i)

$\rm \: a + 6d - (a + 2d) = 9 - 5$

$\rm \: a + 6d - a - 2d = 4$

$\rm \: 4d = 4$

$\rm \: d = 1$

Now put the of d on (i)eq

$\rm \: a \: + 2d \: = 5$

$\rm \: a + 2 \times 1 = 5$

$\rm \: a = 5 - 2$

$\rm \: a = 3$

Answer:-

a = 3 , d = 1

Sequences are :- 3 , 4 , 5 , 6 ............