Question

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In an AP 10, 7,4.. -62 . Find 11th
term from last?​

Answer 1

GIVEN :

• $\sf AP : 10 , 7 , 4 , ... , -62$

TO FIND :

• $\sf 11^{th}\:term\: from \:last ?$

SOLUTION :

$\qquad:\implies\:\sf a_{n} = a + (n - 1)d$

★ Values that we have :

• $\sf a_{n} = \bold{-62}$
• $\sf a = \bold{10}$
• $\sf d = 7 - 10 = \bold{-3}$

★ Putting all values :

$\qquad:\implies\:\sf -62 = 10 + (n - 1)(-3)$

$\qquad:\implies\:\sf -62 = 10 + (-3n + 3)$

$\qquad:\implies\:\sf -62 = 10 - 3n + 3$

$\qquad:\implies\:\sf -62 = 13 - 3n$

$\qquad:\implies\:\sf -62 - 13 = -3n$

$\qquad:\implies\:\sf -75 = -3n$

$\qquad:\implies\:\sf \cancel{-}75 = \cancel{-}3n$

$\qquad:\implies\:\sf 3n = 75$

$\qquad:\implies\:\sf n = \dfrac{75}{3}$

$\qquad:\implies\:\sf n = \dfrac{\cancel{75}}{\cancel{3}}$

$\qquad:\implies\:\bold {n = \red{25}}$

$\large\underline{\boxed{\tt{Number\:of\:terms\:=\:\rm\purple{25}}}}$

ㅤㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

Now,

→ 11th term from last = (25 - 11) + 1

→ 11th term from last = 14 + 1

→ 11th term from last = 15

$\therefore\underline{\sf {11^{th}\: term \:from \:last \:is \:15^{th}\:term\: from\: first}}$

ㅤㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

So,

$\qquad:\implies\:\sf a_{15} = a + (15 - 1)d$

$\qquad:\implies\:\sf a_{15} = 10 + (14)(-3)$

$\qquad:\implies\:\sf a_{15} = 10 - 42$

$\qquad:\implies\:\bold{a_{15} = \red{-32}}$

This is the required answer.

$\large\underline{\boxed{\tt{11^{th}\:term\:from\:last\:=\:\rm\purple{-32}}}}$

Answer 2

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

$\large{ \underline {\bf {\orange{Solution}}}}$

$\:$

$\tt \green{Formula \: \: used}$

$\:$

$\begin{gathered}{ \underline{ \boxed{ \bf{ \blue{ a_{n} = a + (n - 1)d }}}}} \end{gathered}$

$\:$

$\tt \green{Given \: \: values }$

$\:$

• $\bf{ a_{n} = - 62}$
• $\bf{a = 10}$
• $\bf{d = 7 - 10 = - 3}$

$\:$

$\qquad : \rightarrow \sf \blue{ - 62 = 10 + (n - 1)( - 3)}$

$\:$

$\qquad : \rightarrow \bf \blue{ - 62 = 10 + ( - 3n + 3)}$

$\:$

$\qquad : \rightarrow \bf \blue{ - 62 = 13 - 3n}$

$\:$

$\qquad : \rightarrow \bf \blue{ - 62 - 13 = - 3n}$

$\:$

$\qquad : \rightarrow \bf \blue{ - 75 = - 3n}$

$\:$

$\qquad : \rightarrow \bf \blue{ \cancel - 75 = \cancel - 3n}$

$\:$

$\qquad : \rightarrow \bf \blue{3n = 75}$

$\:$

$\qquad : \rightarrow \bf \blue{n = \dfrac{75}{3} }$

$\:$

$\qquad : \rightarrow \bf \blue{n = \cancel{ \dfrac{75}{3}} }$

$\:$

$\qquad : \rightarrow \bf \blue{n = 25}$

$\:$

$\tt \green{Number \: of \: terms = 25}$

$\:$

Now,

$\:$

$\bf{ {11}^{th} \: term \: from \: last \: = (25 - 11) + 1} \\ \bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 14 + 1} \\ \bf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 15}$

$\:$

$\tt \purple{.°. \: {11}^{th} \: \: term \: from \: last \: is \: {15}^{th} \: term \: from \: first.}$

$\:$

$\:$

$\quad \quad : \implies \bf \orange{ a_{15} = a + (15 - 1)d}$

$\:$

$\quad : \implies \bf \orange{ a_{15} = 10 + (14) + ( - 3) }$

$\:$

$\qquad : \implies \bf \orange{ a_{15} = 10 - 42}$

$\:$

$\qquad : \implies \bf \orange{ a_{15} = - 32 }$

$\:$

$\tt \green{Required \: \: a nswer }$

$\:$

${\underline { \boxed{ \sf{ \pink{ {11}^{th}term \: from \: last \: = - 32}}}}}$

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

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