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In an A.P. (with usual notations) : (iii) given d = \[5\], \[{{\mathbf{S}}_{\mathbf{9}}}~=\text{ }\mathbf{75}\], find a and a9. (iv) given \[\mathbf{a}\text{ }=\text{ }\mathbf{8},\text{ }{{\mathbf{a}}_{\mathbf{n}}}~=\text{ }\mathbf{62},\text{ }{{\mathbf{S}}_{\mathbf{n}}}~=\text{ }\mathbf{210}\], find n and d
In an A.P. (with usual notations) : (iii) given d = \[5\], \[{{\mathbf{S}}_{\mathbf{9}}}~=\text{ }\mathbf{75}\], find a and a9. (iv) given \[\mathbf{a}\text{ }=\text{ }\mathbf{8},\text{ }{{\mathbf{a}}_{\mathbf{n}}}~=\text{ }\mathbf{62},\text{ }{{\mathbf{S}}_{\mathbf{n}}}~=\text{ }\mathbf{210}\], find n and d
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
In an A.P. (with usual notations) : (iii) given d = \[5\], \[{{\mathbf{S}}_{\mathbf{9}}}~=\text{ }\mathbf{75}\], find a and a9. (iv) given \[\mathbf{a}\text{ }=\text{ }\mathbf{8},\text{ }{{\mathbf{a}}_{\mathbf{n}}}~=\text{ }\mathbf{62},\text{ }{{\mathbf{S}}_{\mathbf{n}}}~=\text{ }\mathbf{210}\], find n and d

From the question it is given that,

Common difference d = \[5\]

\[{{\mathbf{S}}_{\mathbf{9}}}~=\text{ }\mathbf{75}\]

We know that, an = a + (n – 1)d

\[\begin{array}{*{35}{l}}

{{a}_{9}}~=\text{ }a\text{ }+\text{ }\left( 9\text{ }\text{ }1 \right)5  \\

{{a}_{9}}~=\text{ }a\text{ }+\text{ }45\text{ }\text{ }5  \\

\end{array}\]

\[{{a}_{9}}~=\text{ }a\text{ }+\text{ }40\]… [equation (i)]

Then, \[{{S}_{9}}~=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right)\]

\[\begin{array}{*{35}{l}}

75\text{ }=\text{ }\left( 9/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( 9\text{ }\text{ }1 \right)5 \right)  \\

75\text{ }=\text{ }\left( 9/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( 8 \right)5 \right)  \\

\left( 75\text{ }\times \text{ }2 \right)/9\text{ }=\text{ }2a\text{ }+\text{ }40  \\

150/9\text{ }=\text{ }2a\text{ }+\text{ }40  \\

2a\text{ }=\text{ }150/9\text{ }\text{ }40  \\

2a\text{ }=\text{ }50/3\text{ }\text{ }40  \\

2a\text{ }=\text{ }\left( 50\text{ }\text{ }120 \right)/3  \\

2a\text{ }=\text{ }-70/3  \\

a\text{ }=\text{ }-70/\left( 3\text{ }\times \text{ }2 \right)  \\

a\text{ }=\text{ }\text{ }35/3  \\

\end{array}\]

Now, substitute the value of a in equation (i),

\[\begin{array}{*{35}{l}}

   {{a}_{9}}~=a\text{ }+\text{ }40  \\

   =\text{ }-35/3\text{ }+\text{ }40  \\

   =\text{ }\left( -35\text{ }+\text{ }120 \right)/3  \\

   =\text{ }85/3  \\

\end{array}\]


(iv) given \[\mathbf{a}\text{ }=\text{ }\mathbf{8},\text{ }{{\mathbf{a}}_{\mathbf{n}}}~=\text{ }\mathbf{62},\text{ }{{\mathbf{S}}_{\mathbf{n}}}~=\text{ }\mathbf{210}\], find n and d

Solution:-

From the question it is give that,

First term a = \[8\],

\[{{a}_{n~}}=\text{ }62\text{ }and\text{ }{{S}_{n}}~=\text{ }210\]

We know that, \[{{a}_{n~}}\] = a + (n – 1)d

\[\begin{array}{*{35}{l}}

62\text{ }=\text{ }8\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d  \\

\left( n\text{ }\text{ }1 \right)d\text{ }=\text{ }62\text{ }\text{ }8  \\

\end{array}\]

\[\left( n\text{ }\text{ }1 \right)d\text{ }=\text{ }54\] … [equation (i)]

Then, \[{{S}_{n}}~=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right)\]

\[210\text{ }=\text{ }\left( n/2 \right)\text{ }\left( \left( 2\text{ }\times \text{ }8 \right)\text{ }+\text{ }54 \right)\ldots \][from equation (i) (n – 1)d = \[54\]]

\[\begin{array}{*{35}{l}}

210\text{ }=\text{ }\left( n/2 \right)\text{ }\left( 16\text{ }+\text{ }54 \right)  \\

420\text{ }=\text{ }n\left( 70 \right)  \\

n\text{ }=\text{ }420/70  \\

n\text{ }=\text{ }6  \\

\end{array}\]

Now, substitute the value of n in equation (i),

\[\begin{array}{*{35}{l}}

\left( n\text{ }\text{ }1 \right)d\text{ }=\text{ }54  \\

\left( 6\text{ }\text{ }1 \right)d\text{ }=\text{ }54  \\

5d\text{ }=\text{ }54  \\

d\text{ }=\text{ }54/5  \\

\end{array}\]

Therefore, d = \[54/5\] and n = \[6\]

i

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