Question

which term of the A.P. 5,9,13,... will be 80 less than its 51st term.

Answer 1

$\large\underline{\sf{Solution-}}$

Given AP series is

$\rm :\longmapsto\:5,9,13, - - - - -$

So, Here,

First term, a = 5

Common difference, d = 9 - 5 = 4

Let we assume that aₙ is 80 less than 51 st term.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

According to statement,

$\red{\rm :\longmapsto\:a_n \: = \: a_{51} - 80}$

$\rm :\longmapsto\:a + (n - 1)d = a + (51 - 1)d - 80$

$\rm :\longmapsto\: (n - 1)d = 50d - 80$

On substituting the value of d = 4, we get

$\rm :\longmapsto\: (n - 1)4 = 50 \times 4 - 80$

$\rm :\longmapsto\: 4(n - 1)= 200 - 80$

$\rm :\longmapsto\: 4(n - 1)= 120$

$\rm :\longmapsto\: n - 1= 30$

$\rm \implies\:\boxed{ \tt{ \: n \: = \: 31 \: }}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: {31}^{st} \: term \: is \: 80 \: less \: than \: {51}^{st} \: term \: }}$

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Explore More

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Answer 2

$\large\underline{\sf\red{Solution-}}$

Given AP series is

$\blue{\rm :\longmapsto\:5,9,13, - - - - -}$

So, Here,

First term, a = 5

Common difference, d = 9 - 5 = 4

Let we assume that aₙ is 80 less than 51 st term.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}\end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.
• a is the first term of the sequence.
• n is the no. of terms.
• d is the common difference.

Tʜᴜs,

According to statement,

$\red{\rm :\longmapsto\:a_n \: = \: a_{51} - 80}$

$\blue{\rm :\longmapsto\:a + (n - 1)d = a + (51 - 1)d - 80}$

$\orange{\rm :\longmapsto\: (n - 1)d = 50d - 80}$

On substituting the value of d = 4, we get

$\green{\rm :\longmapsto\: (n - 1)4 = 50 \times 4 - 80}$

$\purple{\rm :\longmapsto\: 4(n - 1)= 200 - 80}$

$\red{\rm :\longmapsto\: 4(n - 1)= 120}$

$\blue{\rm :\longmapsto\: n - 1= 30}$

$\pink{\rm \implies\:\boxed{ \tt{ \: n \: = \: 31 \: }}}$

Hence,

$\orange{\rm \implies\:\boxed{ \tt{ \: {31}^{st} \: term \: is \: 80 \: less \: than \: {51}^{st} \: term \: }}}$

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↝ Sum of n terms of an arithmetic sequence is,

$\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}\end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.
• a is the first term of the sequence.
• n is the no. of terms.
• d is the common difference.

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