Question

if the numbers n-2,4n-1 and 5n+2 are in A.P find the value of n

Answer 1

Answer:

Step-by-step explanation:

n−2,4n−1,5n+2 are in A.P.

First term, a  1=n−2

Second term, a  2

​  

=4n−1

Third term, a  

3

​  

=5n+2

Common difference (difference of two consecutive terms),  

d=a  

2

​  

−a  

1

​  

=a  

3

​  

−a  

2

​  

 

4n−1−(n−2)=5n+2−(4n−1)

4n−1−n+2=5n+2−4n+1

3n+1=n+3

2n=2

n=1

So,  

a  

1

​  

=n−2=1−2=−1

a  

2

​  

=4n−1=4−1=3

and d=a  

2

​  

−a  

1

​  

=3−(−1)=4

nth term of an AP is ,

a  

n

​  

=a+(n−1)d

where,

a=first term

d=common difference

Therefore,

a  

4

​  

=a+3d=−1+12=11

a  

5

​  

=a+4d=−1+16=15

Answer 2

$\bf\huge\blue{\underline{\underline{ Question : }}}$

If the numbers n-2,4n-1 and 5n+2 are in A.P find the value of n.

$\bf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• The numbers n-2,4n-1 and 5n+2 are in A.P

★ To find,

• Value of n.

★ Let,

• a1 = n - 2
• a2 = 4n - 1
• a3 = 5n + 2

We know that,

➡ a2 - a1 = a3 - a2

$\sf \implies (4n - 1) -(n-2) = (5n+2)-(4n-1)$

$\sf \implies 4n-1-n+2 = 5n+2-4n+1$

$\sf \implies 3n+1 = n+3$

$\sf \implies 3n-n=3-1$

$\sf \implies 2n=2$

$\sf \implies n=\cancel{\cfrac{2}{2}}$

$\sf \implies n=1$

$\underline{\boxed{\rm{\purple{\therefore Hence,\:the\:value\:of\:n=1.}}}}\:\orange{\bigstar}$

★ More Information :

$\boxed{\begin{minipage}{5 cm} AP Formulae \\ \\ : \implies a_{n} = a + (n - 1)d \\ \\ :\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] \end{minipage}}$

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