Question

The 5th and 15th terms of an A.P. are -13 and -17 respectively. Find the sum of first 21 terms of the A.P.

Answer 1

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

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$\green{\underline \bold{Given :}}$

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• 5th term is -13

• 15th term is -17

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$\red{\underline \bold{To \: Find:}}$

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• Sum of first 21 term.

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

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$\underline{\bold{\texttt{We know that :}}}$

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$\purple{\longrightarrow}a_n = a + (n - 1)d$

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• a = First term

• n = number of term

• d = common difference

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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$\sf \longmapsto -13 = a + (5 - 1)d$

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$\bf\dashrightarrow -13 = a + 4d$ -------(1)

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$\sf \longmapsto -17 = a + (15 - 1)d$

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$\sf \longmapsto -17 = a + 14d$

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$\bf \dashrightarrow -a - 14d = 17$ -------(2)

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$\underline{\bold{\texttt{Adding (1) and (2)}}}$

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$\sf \longmapsto -10d = 4$

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$\sf \longmapsto d = \dfrac { -4 } { 10 }$

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$\bf \dashrightarrow d = \dfrac { -2 } { 5 }$

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$\underline{\bold{\texttt{Putting the value of d in (1)}}}$

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$\sf \longmapsto -13 = a + 4(\dfrac { -2 } { 5 } )$

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$\sf \longmapsto -13 = a - \dfrac { 8 } { 5 }$

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$\sf \longmapsto a = -13 + \dfrac { 8 } { 5 }$

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$\sf \longmapsto a = \dfrac { -65 + 8 } { 5 }$

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$\bf \dashrightarrow a = \dfrac { -57 } { 5 }$

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$\underline{\bold{\texttt{Sum of first 21 term:}}}$

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$\purple\longrightarrow$ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

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$\sf \longmapsto S_n = \dfrac { 21} { 2 } (2\times\dfrac { -57 } { 5 } + (21 - 1)\dfrac { -2 } { 5 } )$

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$\sf \longmapsto S_n = \dfrac { 21} { 2 } (\dfrac { -114 } { 5 } - 8)$

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$\sf \longmapsto S_n = \dfrac { 21 } { 2 } ( \dfrac { -114 - 40 } { 5 })$

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$\sf \longmapsto S_n = \dfrac { 21 } { 2 } (\dfrac { -154 } { 5 } )$

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$\sf \longmapsto S_n = \dfrac { -1617 } { 5 }$

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Hence sum of first 21 term is $\dfrac { -1617 } { 5 }$

$\rule{200}5$