Question

The end term of a progression is 3 and minus 8 is the pattern of numbers so formed in AP if so find its 16th term

Answer 1

Original question

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The nth term of a progression is 3n-8. is the pattern of number so formed in ap. if so find the 16th term?

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$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• nth term of a progression is 3n-8

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$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• The 16th term

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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• First term = a

• Common difference = d

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➠ 3n-8 ------ (1)

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As first term has n = 1

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So,

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〚 Now putting n = 1 in (1) 〛

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$\underline{\bold{\texttt{First term :}}}$

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➠ 3n - 8

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➜ 3(1) - 8

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➜ 3 - 8

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➜ $\sf a_1$ = -5 = a

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As second term has n = 2

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So,

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〚 Now putting n = 2 in (1) 〛

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$\underline{\bold{\texttt{Second term :}}}$

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➜ 3n - 8

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➜ 3(2) - 8

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➜ 6 - 8

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➜$\sf a_2$ = -2

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$\underline{\bold{\texttt{Common difference :}}}$

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➠ $\sf a_2 - a_1 = d$

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➜ -2 - (-5)

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➜ -2 + 5

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➜ d = 3

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$\underline{\bold{\texttt{For nth term -}}}$

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➠ $\sf a_n = a + (n - 1)d$ ----- (2)

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$\underline{\bold{\texttt{For 16th term :}}}$

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• a = -5

• d = 3

• n = 16

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Putting these values in (2)

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➜ $\sf a_{16} = -5 + (16 - 1)3$

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➜ $\sf a_{16} = -5 + (15)3$

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➜ $\sf a_{16} = -5 + 45$

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➨ $\sf a_{16} = 40$

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• Hence 16th term of the given AP is 40

Answer 2

Original question

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• The nth term of a progression is 3n-8. is the pattern of number so formed in ap. if so find the 16th term?

$\:\:$

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• nth term of a progression is 3n-8

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

• The 16th term

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

First term = a

Common difference = d

$\:\:$

➠ 3n-8 ------ (1)

$\:\:$

As first term has n = 1

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So,

$\:\:$

〚 Now putting n = 1 in (1) 〛

$\:\:$

$\underline{\bold{\texttt{First term :}}}$

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➠ 3n - 8

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➜ 3(1) - 8

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➜ 3 - 8

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➜ $\sf a_1$ = -5 = a

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As second term has n = 2

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So,

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〚 Now putting n = 2 in (1) 〛

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$\underline{\bold{\texttt{Second term :}}}$

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➜ 3n - 8

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➜ 3(2) - 8

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➜ 6 - 8

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➜$\sf a_2$ = -2

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$\underline{\bold{\texttt{Common difference :}}}$

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➠ $\sf a_2 - a_1 = d$

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➜ -2 - (-5)

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➜ -2 + 5

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➜ d = 3

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$\underline{\bold{\texttt{For nth term -}}}$

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➠ $\sf a_n = a + (n - 1)d$ ----- (2)

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$\underline{\bold{\texttt{For 16th term :}}}$

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a = -5

d = 3

n = 16

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Putting these values in (2)

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➜ $\sf a_{16} = -5 + (16 - 1)3$

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➜ $\sf a_{16} = -5 + (15)3$

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➜ $\sf a_{16} = -5 + 45$

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➨ $\sf a_{16} = 40$

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• Hence 16th term of the given AP is 40