Find the indicated terms in each of following A.P.s: (i) \[\mathbf{1},\text{ }\mathbf{6},\text{ }\mathbf{11},\text{ }\mathbf{16},\text{ }\ldots ;\text{ }{{\mathbf{a}}_{\mathbf{20}}}\] (ii) \[-\mathbf{4},\text{ }-\mathbf{7},\text{ }-\mathbf{10},\text{ }-\mathbf{13},\text{ }\ldots ,\text{ }{{\mathbf{a}}_{\mathbf{25}}},\text{ }{{\mathbf{a}}_{\mathbf{n}}}\]

Find the indicated terms in each of following A.P.s: (i)

$\mathbf{1},\text{ }\mathbf{6},\text{ }\mathbf{11},\text{ }\mathbf{16},\text{ }\ldots ;\text{ }{{\mathbf{a}}_{\mathbf{20}}}$

(ii)

$-\mathbf{4},\text{ }-\mathbf{7},\text{ }-\mathbf{10},\text{ }-\mathbf{13},\text{ }\ldots ,\text{ }{{\mathbf{a}}_{\mathbf{25}}},\text{ }{{\mathbf{a}}_{\mathbf{n}}}$

Find the indicated terms in each of following A.P.s: (i)

$\mathbf{1},\text{ }\mathbf{6},\text{ }\mathbf{11},\text{ }\mathbf{16},\text{ }\ldots ;\text{ }{{\mathbf{a}}_{\mathbf{20}}}$

(ii)

$-\mathbf{4},\text{ }-\mathbf{7},\text{ }-\mathbf{10},\text{ }-\mathbf{13},\text{ }\ldots ,\text{ }{{\mathbf{a}}_{\mathbf{25}}},\text{ }{{\mathbf{a}}_{\mathbf{n}}}$

From the question,

The first term a =

$1$

Then, difference d =

$6\text{ }\text{ }1\text{ }=\text{ }5$

$\begin{array}{*{35}{l}} 11\text{ }\text{ }6\text{ }=\text{ }5 \\ 16\text{ }\text{ }11\text{ }=\text{ }5 \\ \end{array}$

Therefore, common difference d =

$5$

From the formula,

${{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d$

So,

$\begin{array}{*{35}{l}} {{a}_{20}}~=\text{ }a\text{ }+\text{ }\left( 20\text{ }\text{ }1 \right)d \\ =\text{ }1\text{ }+\text{ }\left( 20\text{ }\text{ }1 \right)5 \\ =\text{ }1\text{ }+\text{ }\left( 19 \right)5 \\ =\text{ }1\text{ }+\text{ }95 \\ =\text{ }96 \\ \end{array}$

Therefore,

${{a}_{20}}~=\text{ }96$

(ii)

$-\mathbf{4},\text{ }-\mathbf{7},\text{ }-\mathbf{10},\text{ }-\mathbf{13},\text{ }\ldots ,\text{ }{{\mathbf{a}}_{\mathbf{25}}},\text{ }{{\mathbf{a}}_{\mathbf{n}}}$

Solution:-

From the question,

The first term a =

$-4$

Then, difference d =

$-7\text{ }\text{ }\left( -4 \right)\text{ }=\text{ }\text{ }7\text{ }+\text{ }4\text{ }=\text{ }-3$

$\begin{array}{*{35}{l}} -10\text{ }\text{ }\left( -7 \right)\text{ }=\text{ }-10\text{ }+\text{ }7\text{ }=\text{ }-3 \\ -13\text{ }\text{ }\left( -10 \right)\text{ }=\text{ }-13\text{ }+\text{ }10\text{ }=\text{ }-3 \\ \end{array}$

Therefore, common difference d =

$~-3$

From the formula,

${{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d$

So,

$\begin{array}{*{35}{l}} {{a}_{25}}~=\text{ }a\text{ }+\text{ }\left( 25\text{ }\text{ }1 \right)d \\ =\text{ }-4\text{ }+\text{ }\left( 25\text{ }\text{ }1 \right)\left( -3 \right) \\ =\text{ }-4\text{ }+\text{ }\left( 24 \right)-3 \\ =\text{ }\text{ }4\text{ }\text{ }72 \\ =\text{ }-76 \\ \end{array}$

Therefore,

${{a}_{25}}~=\text{ }-76$

Now,

$\begin{array}{*{35}{l}} {{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ {{a}_{n}}~=\text{ }-4\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)-3 \\ =\text{ }-4\text{ }\text{ }3n\text{ }+\text{ }3 \\ =\text{ }-1\text{ }\text{ }3n \\ \end{array}$

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