Question

The fourth and fifth terms of an arithmetic progression are 45 and 50 respectively find (a)the common difference (b)the first term and (c)the tenth term

Answer 1

Question :-

The fourth and fifth terms of an arithmetic progession are 45 and 50 respectively . Find ( a ) the common difference ( b ) the first term and ( c) then tenth term

Answer :-

Given :-

Fourth and Fifth terms of the arithmetic progession are 45 & 50 .

Required to find :-

• The common difference

• The first term

• The tenth term

Formula used :-

$\large{\leadsto{\boxed{\tt{ {a}_{nth} = a + ( n - 1 ) d }}}}$

Solution :-

Given that :-

Fourth term is 45 and Fifth term is 50 .

So,

He asked us to find the common difference , first term and the tenth term .

So,

4th term

= a + 3d = 45 $\longrightarrow{\tt{Equation \; 1 }}$

consider this as equation 1

5th term

= a + 4d = 50 $\longrightarrow{\tt{Equation \; 2 }}$

consider this as equation 2

So,

Now solve this two equation simultaneously by using eliminating method

Hence,

Subtract equation 1 from equation 2

So,

a + 4d = 50

a + 3d = 45

(-) (-) (-)

-----------------

0 + d = 5

Hence,

d = 5

This implies,

Common difference (d) = 5

However,

Substitute this " d " value in equation 1

So,

a + 3d = 45

a + 3 ( 5 ) = 45

a + 15 = 45

a = 45 - 15

a = 30

Hence,

a = 30

This implies ,

The first term ( a ) = 30

However,

using the formula we can find the 10th term .

The formula is ,

$\large{\leadsto{\boxed{\tt{ {a}_{nth} = a + ( n - 1 ) d }}}}$

Here,

a = first term

d = common difference

n = the term which you want to find

So,

By substituting the required values

we get,

$\longrightarrow{\tt{ {a}_{nth} = {a}_{10}}}$

$\longrightarrow{\tt{ {a}_{10} = 30 + ( 10 - 1 ) 5 }}$

$\longrightarrow{\tt{ {a}_{10} = 30 + (9) 5 }}$

$\longrightarrow{\tt{ {a}_{10} = 30 + 45 }}$

$\longrightarrow{\tt{ {a}_{10} = 75 }}$

$\large{\leadsto{\boxed{\tt{ 10th \; term \; = 75 }}}}$

Points to remember :-

1. Formula to find the nth term is ,

$\large{\leadsto{\boxed{\tt{ {a}_{nth} = a + ( n - 1 ) d }}}}$

2. Formula to find the sum of nth term is ,

$\large{\leadsto{\boxed{\tt{ {s}_{nth} = \dfrac{n}{2}[2a + ( n - 1 ) d ]}}}}$

The simple form of this above formula is,

$\large{\leadsto{\boxed{\tt{ {s}_{nth} = \dfrac{n}{2}[First \; term + Last \; term ]}}}}$

3. The arithmetic progession of the above question is

a = 30 , 35 , 40 , - - - - - - - - , 75

( This is for your additional knowledge )

Answer 2

Answer:

Step-by-step explanation:

Given :-

The fourth and fifth terms of an arithmetic progression are 45 and 50.

To Find :-

The common difference, d = ??

The first term, a = ??

the tenth term = ??

Formula to be used :-

a(n) = a + (n - 1)d

Solution :-

Let a be the first term and d be the common difference of given A.P.

Here, we have

a + 3d = 45 ... (i)

And, a + 4d = 50 .... (ii)

By solving Eq (i) and (ii), we get

 a + 3d = 45

 a + 4d = 50

-    +        +

___________

⇒ d = 5

Pudding d's value in Eq (i), we get

⇒ a + 3d = 45

⇒ a + 3 × 5 = 45

⇒ a + 15 = 45

⇒ a = 45 - 15

⇒ a = 30

Hence, the first term is 30 and the common difference is 5.

Now, tenth term

We know that,

a(n) = a + (n - 1)d

⇒ a(10) = 30 + (10 - 1)5

⇒ a(10) = 30 + 9 × 5

⇒ a(10) = 30 + 45

⇒ a(10) = 75

Hence, the tenth term of an A.P. is 75.