Question

The algebraic form of an arithematic sequence is5n-3.find the sum of the first 20th term

Answer 1

$\mathbb{\huge{\fbox{\green{AnSwEr:-}}}}$

✏ The sum of the first 20th term of AP is 990.

$\mathcal{\huge{\fbox{\red{Formulae\:Used:-}}}}$

✨ $\mathcal{\small{\fbox{\orange{S\:=\:n\:/\:2(2a \:+ \:(n \:-\: 1)d}}}}$

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

Given ,

nth term of an AP is 5n - 3

Thus ,

First term = 5(1) - 3 = 5 - 3 = 2

Second term = 5(2) - 3 = 10 - 3 = 7

The common difference is (7-2) = 5

We know that , the sum of first nth term of an AP is given by

S = n/2(2a + (n - 1)d

Substitute the known values , we get

▶ S = 20/2 (2×2+(20-1)5)

▶ S = 10 (4+ 19× 5)

▶ S = 10(4 + 95)

▶ S = 10 × 99

▶S = 990

Therefore, the sum of the first 20th term is 990.

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Answer 2

Question : -

The algebraic form of the nth term of arithmetic sequence is 5n - 3 . Find the sum of the first 20 terms ?

Answer : -

Given : -

The algebraic form of the nth term of arithmetic sequence is 5n - 3 .

Required to find : -

• Sum of first 20 terms ?

Formula used : -

$\large{\boxed{\tt{ {S}_{nth} = \dfrac{ n}{ 2} [ first \ term + last \ term ] }}}$

or

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

Here,

a = first term

d = common difference

n = term number

Solution : -

The algebraic form of the nth term of arithmetic sequence is 5n - 3 .

So,

T ( nth ) = 5n - 3

Now,

Let's find the AP .

1st term ;

T ( 1 ) =

=> 5 ( 1 ) - 3

=> 5 - 3

=> 2

2nd term ;

T ( 2 ) =

=> 5 ( 2 ) - 3

=> 10 - 3

=> 7

3rd term ;

T ( 3 ) =

=> 5 ( 3 ) - 3

=> 15 - 3

=> 12

4th term ;

T ( 4 ) =

=> 5 ( 4 ) - 3

=> 20 - 3

=> 17

Hence,

AP = 2 , 7 , 12 , 17 , . . . . . . .

Now,

Let's find the value of a , d

Since,

a = first term

• a = 2

Common difference =

=> ( 2nd term - 1st term ) = ( 3rd term - 2nd term )

=> ( 7 - 2 ) = ( 12 - 7 )

=> ( 5 ) = ( 5 )

• Common difference ( d ) = 5

Now,

Let's find the 20th term because 20th term is the last term till which we need to find the sum .

So,

20th term can be written as ,

a + 19d ( or ) a + ( 20 - 1 ) d

Substitute the values of a , d ;

=> 2 + 19 ( 5 )

=> 2 + 95

=> 97

Hence,

• 20th term = 97

Using the formula ;

$\large{\boxed{\tt{ {S}_{nth} = \dfrac{ n}{ 2} [ first \ term + last \ term ] }}}$

This implies ;

$\longrightarrow \tt{ { S}_{nth} = {S}_{20} } \\ \\ \longrightarrow \tt{ { S}_{20 } = \dfrac{ 20}{2} [ 2 + 97 ] } \\ \\ \longrightarrow \tt{ { S }_{20} = \dfrac{ 20}{2} [ 99 ] } \\ \\ \longrightarrow \tt{ {S}_{20} = 10 [ 99 ] } \\ \\\longrightarrow \tt{ {S}_{20} = 10 \times 99 } \\ \\ \longrightarrow \tt{ {S}_{20} = 990 }$

Therefore,

Sum of first 20 terms = 990

Additional Information : -

What is an arithmetic progression ?

An arithmetic progression is a sequence of numbers which have a common difference between them . The difference between the terms of the arithmetic progression is constant .

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Difference between Arithmetic progression and a sequence / series ?

An arithmetic progression is a sequence in which the difference the term si constant .

whereas,

A sequence/series a set of numbers which have a common pattern but not difference .

Every arithmetic progression can be taken as a series / sequence .

But,

Every series / sequence can't be considered as an Arithmetic progression .