Question

Determine the AP whose 3rd term is 5 & 7th term is 9

Answer 1

Answer:

ATQ

a3 = 5 a7 = 9

Therefore

a + 2d = 5 Eq1

a + 6d = 9 Eq2

Using Elimination Method

a + 6d = 9

-a -2d = -5

4d = 4

Therefore d = 1

Putting value of d in Eq 1

a + 2(1) = 5

a + 2 = 5

a = 3

Now

a = 3

a2 = a + d = 3 + 1 = 4

a3 = 5

Therefore AP is 3,4,5,--------

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Answer 2

Given :-

3rd term = 5

7th term = 9

Required to find :-

• The arithmetic progession

Solution :-

Given information :-

3rd term = 5

7th term = 9

we need to find the arithmetic progession

Consider the given information ;

The 3rd term is 5 and 7th term is 9

But,

The 3rd term is represented as " a + 2d "

So,

• a + 2d = 5 $\longrightarrow{\text{\pink{Equation - 1 }}}$

consider this as equation 1

Similarly,

The 7th term is represented as " a + 6d "

So,

• a + 6d = 9 $\longrightarrow{\text{\pink{Equation - 2 }}}$

consider this as equation 2

Now,

We need to solve these two equations simultaneously ;

Using elimination method ;

>> Subtract equation 1 from equation 2 <<

$\large \tt \: \: \: a + 6d = 9 \\ \large \tt \: \: \: a + 2d = 5 \\ \large \tt \underline{( - ) ( - ) \: \: \: ( - ) \: } \\ \large \tt \underline{ \: \: \: \: \: \: \: + 4d = 4 \: \: \: } \\ \large \implies \tt d = \dfrac{4}{4} \\ \\ \large \tt \implies d = 1$

So,

• Common difference ( d ) = 1

Now substitute the value of d in equation 1

➟ a + 2d = 5

➟ a + 2 ( 1 ) = 5

➟ a + 2 = 5

➟ a = 5 - 2

➟ a = 3

However,

• First term ( a ) = 3

Hence,

Let's form the arithmetic progession using the format of the arithmetic progession

The format of the arithmetic progession is ;

AP = a , a + d , a + 2d , a + 3d , a + 4d . . . . . .

So,

The required AP is ;

AP = 3 , 3 + 1 , 3 + 2(1) , 3 + 3(1) , 3 + 4(1) . . . . .

AP = 3 , 4 , 5 , 6 , 7 . . . . . .

Concept used :-

1. Formulae used in arithmetic progession

$\large{\dagger{\rm{ {a}_{nth} = a + ( n - 1 ) d }}}$

$\large{\dagger{\rm{ {S}_{nth} = \dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}}$

2. To find the common difference of any given arithmetic progession we need to apply a small trick

That is ,

Common difference = ( 2nd term - 1st term ) = ( 3rd term - 2nd term )

Note :- If the common difference of the any sequence is constant then only it is said to be as Arithmetic progession