Question

find the value of n if

1+4+9+10+_ _ _ _ _ _ _ upto n term = 287​

Answer 1

$\underbrace{\underline{\bf{\bigstar\:Required~Answer:-}}}$

» We are given with the sum of n terms of an Arithmetic Progression.

» We would first find the first term and the common difference of the AP, then we would put the value of of the first term and the common difference in the formula used for finding the sum of terms of AP.

» From there, we would get the value of "n". After finding n, we would use the formula to find the n th term of an AP so as to get the answer.

◐ Finding the First term and the common difference :

Given AP :

➡ $\quad$ 1 , 4 , 7 , 10 , ... , n

So,

$\leadsto \boxed{ \sf first \: term \: \bf{(a) = 1}}$

And

$\mapsto \sf \: common \: difference\: ( \bf{a}) = 4 - 1 \\ \\ \leadsto \boxed{ \bf{d= 3}}$

◐ Finding the value of n :

We have :

$\quad\;$ $\bullet$ a = 1

$\quad\;$ $\bullet$ d = 3

$\quad\;$ $\bullet$ Sum (n terms) = 287

We know :

$\odot \: \boxed{ \sf{s _{ \: ( x )} = \frac{x}{2} \big \{2a + \big(x - 1 \big)d \big \}}}$

» Here "x" represents the number of terms.

So,

$\displaystyle \colon \longrightarrow \sf \: 287 = \sf \dfrac{ \bf n}{2} \bigg(2 \times (1) + (n - 1)(3) \bigg)$

$\colon \longrightarrow \sf \: 287 \times 2 = \bf{n} \sf \big(2 + 3 \bf{n }- \sf 3)$

$\colon \longrightarrow \sf \: 574 = \bf{n}( \sf3 \bf{n} \sf - 1)$

$\colon \longrightarrow \sf \: 574 = 3 { \bf{n}}^{2} - \bf n$

$\colon \longrightarrow \sf \: 3 { \bf n}^{2} - { \bf{n}} - 574 = 0$

$\colon \longrightarrow \sf \: \underline{3 { \bf{n}}^{2} - 42{ \bf{n}} }+ \underline{41{ \bf{n}} - 574 }= 0$

$\colon \longrightarrow \sf \: 3{ \bf{n}}( { \bf{n}} - 14) + 41( { \bf{n}} - 14) = 0$

$\colon \longrightarrow \sf \: (3{ \bf{n}} + 41)({ \bf{n}} - 14) = 0$

So,

Either :

$\mapsto \sf \: 3n + 41 = 0 \\ \\ \leadsto \sf \: n = \frac{ - 41}{3} \\ \\ \bf \quad \red{rejected}$

Or :

$\mapsto \sf \: n - 14 = 0 \\ \\ \leadsto \boxed{ \green{ \bf n = 14}}$

◐ Finding the n th term :

We have :

$\quad\;$ $\bullet$ a = 1

$\quad\;$ $\bullet$ d = 3

$\quad\;$ $\bullet$ n = 14

We know :

$\odot \: \boxed{ \sf{x {}^{th} \: term \: \big \{a _{ \: (x)} \big \}= a + (x - 1)d}}$

» Here "x" represents the number of terms.

So,

$\colon \implies \sf \: {n}^{th} \: term = a + (n - 1)d \\ \\ \colon \implies \sf \: a _{ \: 14} = 1 + (14 - 1)(3) \\ \\ \colon \implies \sf \: a_{ \: 14} = 1 + 13 \times 3 \\ \\ \colon \implies \sf \: a _{ \: 14} = 1 + 39 \\ \\ \colon \dashrightarrow \boxed{ \boxed{ \purple{ \frak{a _{ \: 14} = 40}}}}$

$\therefore\;\underline{\sf The\:n^{th} \: term \: would \: be \:\pink{\bf 40}.}$

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Answer 2

Answer:

$\underbrace{\underline{\bf{\bigstar\:Required~Answer:-}}}$

» We are given with the sum of n terms of an Arithmetic Progression.

» We would first find the first term and the common difference of the AP, then we would put the value of of the first term and the common difference in the formula used for finding the sum of terms of AP.

» From there, we would get the value of "n". After finding n, we would use the formula to find the n th term of an AP so as to get the answer.

◐ Finding the First term and the common difference :

Given AP :

$➡ \quad 1 , 4 , 7 , 10 , ... , n$

So,

$\leadsto \boxed{ \sf first \: term \: \bf{(a) = 1}}$

And,

$\begin{gathered} \mapsto \sf \: common \: difference\: ( \bf{a}) = 4 - 1 \\ \\ \leadsto \boxed{ \bf{d= 3}}\end{gathered}$

◐ Finding the value of n :

We have,

$\sf \bullet∙ a = 1$

$\sf \bullet∙ d = 3$

$\sf\quad\; \bullet∙ Sum (n terms) = 287$

We knows

$\odot \: \boxed{ \sf{s _{ \: ( x )} = \frac{x}{2} \{2a + \big(x - 1 \big)d \}}}$

» Here "x" represents the number of terms.

So,

$\begin{gathered} \displaystyle \colon \longrightarrow \sf \: 287 = \sf \dfrac{ \bf n}{2} \bigg(2 \times (1) + (n - 1)(3) \bigg) \\ \\ \end{gathered}$

$\begin{gathered} \colon \longrightarrow \sf \: 287 \times 2 = \bf{n} \sf \big(2 + 3 \bf{n }- \sf 3) \\ \\ \end{gathered}$

$\begin{gathered} \colon \longrightarrow \sf \: 574 = \bf{n}( \sf3 \bf{n} \sf - 1) \\ \\ \end{gathered}$

$\begin{gathered} \colon \longrightarrow \sf \: 574 = \bf{n}( \sf3 \bf{n} \sf - 1) \\ \\ \end{gathered}$

$\begin{gathered} \colon \longrightarrow \sf \: 574 = 3 { \bf{n}}^{2} - \bf n \\ \\ \end{gathered}</p><p>$

$\begin{gathered} \colon \longrightarrow \sf \: 3 { \bf n}^{2} - { \bf{n}} - 574 = 0 \\ \\ \end{gathered}$

$\begin{gathered} \colon \longrightarrow \sf \: \underline{3 { \bf{n}}^{2} - 42{ \bf{n}} }+ \underline{41{ \bf{n}} - 574 }= 0 \\ \\ \end{gathered}$

$\begin{gathered} \colon \longrightarrow \sf \: (3{ \bf{n}} + 41)({ \bf{n}} - 14) = 0 \\ \\ \end{gathered}$

So,

Either,

$\begin{gathered} \mapsto \sf \: 3n + 41 = 0 \\ \\ \leadsto \sf \: n = \frac{ - 41}{3} \\ \\ \bf \quad \red{rejected}\end{gathered}$

Or,

$\begin{gathered} \mapsto \sf \: n - 14 = 0 \\ \\ \leadsto \boxed{ \green{ \bf n = 14}}\end{gathered}$

◐ Finding the n th term :

$\quad\; \bullet∙ a = 1$

$\quad\; \bullet∙ d = 3$

$\quad\; \bullet∙ n = 14$

We know

$\odot \: \boxed{ \sf{x {}^{th} \: term \: \{a _{ \: (x)} \}= a + (x - 1)d}}$

» Here "x" represents the number of terms.

So,

$\begin{gathered} \colon \implies \sf \: {n}^{th} \: term = a + (n - 1)d \\ \\ \colon \implies \sf \: a _{ \: 14} = 1 + (14 - 1)(3) \\ \\ \colon \implies \sf \: a_{ \: 14} = 1 + 13 \times 3 \\ \\ \colon \implies \sf \: a _{ \: 14} = 1 + 39 \\ \\ \colon \dashrightarrow \boxed{ \boxed{ \purple{ \frak{a _{ \: 14} = 40}}}}\end{gathered}$

$\begin{gathered}\therefore\;\underline{\sf The\:n^{th} \: term \: would \: be \:\pink{\bf 40}.}\\\end{gathered}$

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