(i) Find the \[20\]th term from the last term of the A.P. \[\mathbf{3},\text{ }\mathbf{8},\text{ }\mathbf{13},\text{ }\ldots ,\text{ }\mathbf{253}\]. (ii) Find the \[{{12}^{th}}\] from the end of the A.P. \[\text{ }\mathbf{2},\text{ }\text{ }\mathbf{4},\text{ }\text{ }\mathbf{6},\text{ }\ldots ,\text{ }\text{ }\mathbf{100}\].

(i) Find the

$20$

th term from the last term of the A.P.

$\mathbf{3},\text{ }\mathbf{8},\text{ }\mathbf{13},\text{ }\ldots ,\text{ }\mathbf{253}$

. (ii) Find the

${{12}^{th}}$

from the end of the A.P.

$\text{ }\mathbf{2},\text{ }\text{ }\mathbf{4},\text{ }\text{ }\mathbf{6},\text{ }\ldots ,\text{ }\text{ }\mathbf{100}$

(i) Find the

$20$

th term from the last term of the A.P.

$\mathbf{3},\text{ }\mathbf{8},\text{ }\mathbf{13},\text{ }\ldots ,\text{ }\mathbf{253}$

. (ii) Find the

${{12}^{th}}$

from the end of the A.P.

$\text{ }\mathbf{2},\text{ }\text{ }\mathbf{4},\text{ }\text{ }\mathbf{6},\text{ }\ldots ,\text{ }\text{ }\mathbf{100}$

Let us assume

$253\text{ }as\text{ }{{n}^{th}}~$

term.

From the question,

The first term a =

$3$

Then, difference d

$\begin{array}{*{35}{l}} =\text{ }8\text{ }\text{ }3\text{ }=\text{ }5 \\ 13\text{ }\text{ }8\text{ }=\text{ }5 \\ \end{array}$

Therefore, common difference d =

$5$

$\begin{array}{*{35}{l}} {{T}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ So,\text{ }253\text{ }=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ 253\text{ }=\text{ }3\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)5 \\ 253\text{ }=\text{ }3\text{ }+\text{ }5n\text{ }\text{ }5 \\ 253\text{ }=\text{ }-2\text{ }+\text{ }5n \\ 5n\text{ }=\text{ }253\text{ }+\text{ }2 \\ 5n\text{ }=\text{ }255 \\ n\text{ }=\text{ }255/5 \\ n\text{ }=\text{ }51 \\ \end{array}$

Therefore,

$253\text{ }is\text{ }{{51}^{th}}~$

term.

Now, assume ‘P’ be the

$20$

th term from the last.

Then,

$\begin{array}{*{35}{l}} P\text{ }=\text{ }L\text{ }\text{ }\left( n\text{ }\text{ }1 \right)d \\ =\text{ }253\text{ }\text{ }\left( 20\text{ }\text{ }1 \right)\text{ }5 \\ =\text{ }253\text{ }\text{ }\left( 19 \right)\text{ }5 \\ =\text{ }253\text{ }\text{ }95 \\ P\text{ }=\text{ }158 \\ \end{array}$

Therefore,

$158\text{ }is\text{ }the\text{ }{{20}^{th}}~$

term from the last.

(ii) Find the

${{12}^{th}}$

from the end of the A.P.

$\text{ }\mathbf{2},\text{ }\text{ }\mathbf{4},\text{ }\text{ }\mathbf{6},\text{ }\ldots ,\text{ }\text{ }\mathbf{100}$

Solution:-

Let us assume

$-100$

as n^th term.

From the question,

The first term a =

$-2$

Then, difference d

$\begin{array}{*{35}{l}} =\text{ }\text{ }4\text{ }\text{ }\left( -2 \right)\text{ }=\text{ }\text{ }4\text{ }+\text{ }2\text{ }=\text{ }-2 \\ -6\text{ }\text{ }\left( -4 \right)\text{ }=\text{ }-6\text{ }+\text{ }4\text{ }=\text{ }\text{ }2 \\ \end{array}$

Therefore, common difference d =

$-2$

$\begin{array}{*{35}{l}} {{T}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ So,\text{ }\text{ }100\text{ }=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ \text{ }100\text{ }=\text{ }-2\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\left( -2 \right) \\ \text{ }100\text{ }=\text{ }-2\text{ }\text{ }2n\text{ }+\text{ }2 \\ \text{ }100\text{ }=\text{ }-2n \\ n\text{ }=\text{ }-100/-2 \\ n\text{ }=\text{ }50 \\ \end{array}$

Therefore,

$-100$

${{50}^{th}}$

term.

Now, assume ‘P’ be the

${{12}^{th}}$

term from the last.

Then,

P = L – (n – 1)d

= -100 – (12 – 1) (-2)

= -100 – (11) (-2)

= -100 + 22

P = – 78

Therefore,

$-78$

is the

${{12}^{th}}$

term from the last of the A.P.

$\text{ }2,\text{ }\text{ }4,\text{ }\text{ }6,\text{ }\ldots$

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