Question

If the sum of terms of the AP 3,7,11,...... is 210, then the number of terms is

Answer 1

Question :-

If the sum of nth terms of the AP 3 , 7 , 11 , . . . . . is 210 . Find the number of terms in it ?

Given :-

Arithmetic Progession ;

AP = 3 , 7 , 11 , . . . . . . . . . .

Sum of the nth term is 210

Required to find :-

• Number of terms in the AP ?

Formula used :-

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

Here,

n = the term number till which you want to find the sum

a = first term

d = common difference

Solution :-

Given data :-

Arithmetic Progession ;

AP = 3 , 7 , 11 , . . . . . . . . . .

Sum of the nth term is 210

we need to find the number of the terms in the AP .

In order to find the number .

Firstly,

we should find the first term , common difference in the AP .

So,

• First term ( a ) = 3

Common difference = ?

This implies ;

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

( 7 - 3 ) = ( 11 - 7 )

( 4 ) = ( 4 )

Hence,

• Common difference ( d ) = 4

Using the formula ;

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

$\tt{ {S}_{nth} = 210 }$

Because , Sum of nth terms is 210

$\tt{ 210 = \dfrac{ n }{ 2 } [ 2 ( 3 ) + ( n - 1 ) 4 ] }$

$\tt{ 210 = \dfrac{ n}{ 2 } [ 6 + ( n - 1 ) 4 ] }$

$\tt{ 210 = \dfrac{ n }{ 2 } [ 6 + 4n - 4 ] }$

$\tt{210 \times 2 == n [ 2 + 4n ] }$

$\tt{ 420 = n \times ( 4n + 2 ) }$

$\tt{ 420 = 4n^2 + 2n }$

$\tt{ 4n^2 + 2n - 420 = 0 }$

$\tt{ 2 ( 2n^2 + n - 210 ) = 0 }$

$\tt{ 2n^2 + n - 210 = \dfrac{0}{2} }$

$\tt{ 2n^2 + n - 210 = 0 }$

$\tt{ 2n^2 + 21n - 20n - 210 = 0 }$

$\tt{ n ( 2n + 21 ) - 10 ( 2n + 21 ) = 0 }$

$\tt{ ( 2n + 21 ) ( n - 10 ) = 0 }$

This implies ;

2n + 21 = 0

2n = -21

n = -21/2

similarly,

n - 10 = 0

n = 10

Since, the number of terms can't be in negative .

Hence,

• Value of n = 10

Therefore ,

Number of terms in the AP = 10

Answer 2

Given ,

First term (a) = 3

Common difference (d) = 4

Sum of first n terms (Sn) = 210

We know that , the sum of first n terms is given by

$\large{\boxed{ \sf{S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}}}$

Thus ,

$\sf \mapsto 210 = \frac{n}{2} \{ 2 \times 3 + (n - 1)4\} \\ \\\sf \mapsto 420 = n(4n + 2) \\ \\\sf \mapsto 4 {(n)}^{2} + 2n - 420 = 0 \\ \\\sf \mapsto 2 {(n)}^{2} + n - 210 = 0$

By middle term splitting formula , we get

$\sf \mapsto 2 {(n)}^{2} - 20n + 21n - 210 = 0 \\ \\\sf \mapsto 2n(n - 10) + 21(n - 10) \\ \\\sf \mapsto (2n + 21)(n - 1) \\ \\\sf \mapsto n = - \frac{21}{2} \: \: or \: \: n= 10$

Since , the number of terms can't be negative and decimal number

$\sf \therefore \underline{The \: required \: number \: of \: terms \: is \: 10}$