Question

the first three terms of an ap are respectively (3n-1),(3n+5) and (5n+1) find n

Please full explain​

Answer 1

$\boxed{ \underline{\underline{\tt { Solution:-}} }} \: \blue✾$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{A.P = (3n-1) \: , (3n+5) \: and \: (5n+1) }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize✾ \: \footnotesize\tt \blue{First \: term \: (a1) = (3n-1) \: }$

$\footnotesize ✾ \: \footnotesize\tt \blue{Second \: term \: (a2) = (3n+5) \: }$

$\footnotesize✾ \: \footnotesize\tt \blue{Third \: term \: (a3) = (5n + 1) \: }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{Note - We \: know \: that \: if \: this \: is \: a \: ap \: then \: their \: common \: difference \: is \: also \: same}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt \blue{So, \: we \: also \: write \: it \: as}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt {a2 - a1 = a3 - a2}$

$\footnotesize \tt {3n + 5 - (3n - 1)= 5n + 1 - (3n + 5)}$

$\footnotesize \tt {3n + 5 - 3n + 1= 5n + 1- 3n - 5}$

$\footnotesize \tt { \cancel3n + 5 - \cancel3n + 1= 2n - 4}$

$\footnotesize \tt { 6= 2n - 4}$

$\footnotesize \tt { - 2n = - 4 - 6}$

$\footnotesize \tt { - 2n = - 10}$

$\footnotesize \tt { \cancel - 2n = \cancel - 10}$

$\tt { n = \frac{10}{2} }$

$\tt { n = \cancel\frac{10}{2} }$

$\boxed{\underline { \underline{\footnotesize \tt { n = 5}}}} \: \footnotesize\blue✾$

Answer 2

Answer:

\boxed{ \underline{\underline{\tt { Solution:-}} }} \: \blue✾

Solution:−

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\: \: \: \: \: \: \: \: \: \: \: \: \: \:

\footnotesize\tt{A.P = (3n-1) \: , (3n+5) \: and \: (5n+1) }A.P=(3n−1),(3n+5)and(5n+1)

\: \: \: \: \: \: \: \: \: \: \: \: \: \:

\footnotesize✾ \: \footnotesize\tt \blue{First \: term \: (a1) = (3n-1) \: }✾Firstterm(a1)=(3n−1)

\footnotesize ✾ \: \footnotesize\tt \blue{Second \: term \: (a2) = (3n+5) \: }✾Secondterm(a2)=(3n+5)

\footnotesize✾ \: \footnotesize\tt \blue{Third \: term \: (a3) = (5n + 1) \: }✾Thirdterm(a3)=(5n+1)

\: \: \: \: \: \: \: \: \: \: \: \: \: \:

\footnotesize\tt{Note - We \: know \: that \: if \: this \: is \: a \: ap \: then \: their \: common \: difference \: is \: also \: same}Note−Weknowthatifthisisaapthentheircommondifferenceisalsosame

\: \: \: \: \: \: \: \: \: \: \: \: \: \:

\footnotesize \tt \blue{So, \: we \: also \: write \: it \: as}So,wealsowriteitas

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:

\footnotesize \tt {a2 - a1 = a3 - a2}a2−a1=a3−a2

\footnotesize \tt {3n + 5 - (3n - 1)= 5n + 1 - (3n + 5)}3n+5−(3n−1)=5n+1−(3n+5)

\footnotesize \tt {3n + 5 - 3n + 1= 5n + 1- 3n - 5}3n+5−3n+1=5n+1−3n−5

\footnotesize \tt { \cancel3n + 5 - \cancel3n + 1= 2n - 4}

3

n+5−

3

n+1=2n−4

\footnotesize \tt { 6= 2n - 4}6=2n−4

\footnotesize \tt { - 2n = - 4 - 6}−2n=−4−6

\footnotesize \tt { - 2n = - 10}−2n=−10

\footnotesize \tt { \cancel - 2n = \cancel - 10}

−

2n=

−

10

\tt { n = \frac{10}{2} }n=

2

10

\tt { n = \cancel\frac{10}{2} }n=

2

10

\boxed{\underline { \underline{\footnotesize \tt { n = 5}}}} \: \footnotesize\blue✾

n=5

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