Question

solve this problem guys by using pen and paper.....​

Answer 1

Question :

In an AP ,

an = 4 , d = 2 , Sn = - 14 , find n and a ?

ANSWER

Given : -

In an AP ,

an = 4 , d = 2 , Sn = - 14 , find n and a ?

Required to find : -

• Find the value of n and a ?

Formula used : -

To find the nth term of any given arithmetic progression is ;

$\sf{ {a}_{nth} = a + (n-1)d }$

To find the sum of any n terms of any given arithmetic progression is ;

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

Solution : -

In an AP ,

an = 4 , d = 2 , Sn = - 14 , find n and a ?

From the given information we can conclude that ;

The nth term of an Arithmetic progression is 4 .

Common difference of the Arithmetic progression is 2 .

The sum of the n terms of the Arithmetic progression is - 14 .

This implies ;

Using the formula , to find the nth term of the AP !

$\sf{ {a}_{nth} = a + ( n - 1 ) d } \\ \\ \sf \to {a}_{nth} = 4 \\ \sf \to d = 2 \\ \\ \sf 4 = a + ( n - 1 )2 \\ \\ \sf 4 = a + 2n - 2 \\ \\ \sf 4 + 2 = a + 2n \\ \\ \sf 6 = a + 2n \\ \\ \sf a + 2n = 6 \longrightarrow \rm{ equation - 1}$

Consider this as equation - 1

Similarly,

Using the formula , to find the sum of n terms of the AP !

$\sf {S}_{nth} = \dfrac{n}{2} [ 2a + ( n - 1 )d ] \\ \\ \sf \to {S}_{nth} = - 14 \\ \sf \to d = 2 \\ \\ \sf - 14 = \dfrac{n}{2} [ 2a + ( n - 1 )2 ] \\ \\ \sf - 14 = \dfrac{n}{2} [ 2a+ 2n - 2 ] \\ \\ \sf - 14 \times 2 = n [ 2a + 2n - 2 ] \\ \\ \sf -28 = n [ 2a + 2n - 2 ] \longrightarrow Equation - 2$

Consider this as Equation 2 .

From Equation 1

a + 2n = 6

a = 6 - 2n

Substitute the value of a in Equation 2

- 28 = n [ 2 ( 6 - 2n ) + 2n - 2 ]

- 28 = n [ 12 - 4n + 2n - 2 ]

- 28 = n [ 12 - 2n - 2 ]

- 28 = 12n - 2n² - 2n

- 28 = 10n - 2n²

2n² - 10n - 28 = 0

2 ( n² - 5n - 14 ) = 0

n² - 5n - 14 = 0/2

n² - 5n - 14 = 0

n² - 7n + 2n - 14 = 0

n ( n - 7 ) + 2 ( n - 7 ) = 0

( n - 7 ) ( n + 2 ) = 0

n = 7 or - 2

since, no. of term s can't be in negative .

value of n = 7

Substitute the value of n in Equation 1

a + 2n = 6

a + 2(7) = 6

a + 14 = 6

a = 6 - 14

a = - 8

value of a = - 8

Therefore,

The value of n = 7 ✔️

Value of a = - 8 ✔️

Answer 2

Question :

In an AP ,

an = 4 , d = 2 , Sn = - 14 , find n and a ?

ANSWER

Given : -

In an AP ,

an = 4 , d = 2 , Sn = - 14 , find n and a ?

Required to find : -

Find the value of n and a ?

Formula used : -

To find the nth term of any given arithmetic progression is ;

= a + (n-1)d }anth​=a+(n−1)d

To find the sum of any n terms of any given arithmetic progression is ;

= [ 2a + (n-1)d ] =2n​[2a+(n−1)d]

Solution : -

In an AP ,

an = 4 , d = 2 , Sn = - 14 , find n and a ?

From the given information we can conclude that ;

The nth term of an Arithmetic progression is 4 .

Common difference of the Arithmetic progression is 2 .

The sum of the n terms of the Arithmetic progression is - 14 .

This implies ;

Using the formula , to find the nth term of the AP !

= a + ( n - 1 ) d }= 4 d
= 2 4 = a + ( n - 1 )24 = a + 2n - 2
4 + 2 = a +2n
6 = a + 2n
a + 2n = 6
⟶equation−1​

Consider this as equation - 1

Similarly,

Using the formula , to find the sum of n terms of the AP !

= [ 2a + ( n - 1 )d ]
= - 14 d = 2 - 1
= [ 2a +( n-1 )2 ] - 14
= [ 2a+ 2n - 2 ] - 14 2
= n [ 2a + 2n - 2 ] -28 = n [ 2a + 2n - 2 ]
Equation - 2 Snth​=2n​[2a+(n−1)d]
→Snth​=−14
→d=2−14=2n​[2a+(n−1)2]−14=2
⟶Equation−2​

Consider this as Equation 2 .

From Equation 1

a + 2n = 6

a = 6 - 2n

Substitute the value of a in Equation 2

- 28 = n [ 2 ( 6 - 2n ) + 2n - 2 ]

- 28 = n [ 12 - 4n + 2n - 2 ]

- 28 = n [ 12 - 2n - 2 ]

- 28 = 12n - 2n² - 2n

- 28 = 10n - 2n²

2n² - 10n - 28 = 0

2 ( n² - 5n - 14 ) = 0

n² - 5n - 14 = 0/2

n² - 5n - 14 = 0

n² - 7n + 2n - 14 = 0

n ( n - 7 ) + 2 ( n - 7 ) = 0

( n - 7 ) ( n + 2 ) = 0

n = 7 or - 2

since, no. of term s can't be in negative .

value of n = 7

Substitute the value of n in Equation 1

a + 2n = 6

a + 2(7) = 6

a + 14 = 6

a = 6 - 14

a = - 8

value of a = - 8

Therefore,

The value of n = 7 

Value of a = - 8