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(i) How many terms of the A.P. \[\mathbf{25},\text{ }\mathbf{22},\text{ }\mathbf{19},\text{ }\ldots \] are needed to give the sum \[116\] ? Also find the last term. (ii) How many terms of the A.P. \[\mathbf{24},\text{ }\mathbf{21},\text{ }\mathbf{18},\text{ }\ldots \] must be taken so that the sum is \[78\] ? Explain the double answer.
(i) How many terms of the A.P. \[\mathbf{25},\text{ }\mathbf{22},\text{ }\mathbf{19},\text{ }\ldots \] are needed to give the sum \[116\] ? Also find the last term. (ii) How many terms of the A.P. \[\mathbf{24},\text{ }\mathbf{21},\text{ }\mathbf{18},\text{ }\ldots \] must be taken so that the sum is \[78\] ? Explain the double answer.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
(i) How many terms of the A.P. \[\mathbf{25},\text{ }\mathbf{22},\text{ }\mathbf{19},\text{ }\ldots \] are needed to give the sum \[116\] ? Also find the last term. (ii) How many terms of the A.P. \[\mathbf{24},\text{ }\mathbf{21},\text{ }\mathbf{18},\text{ }\ldots \] must be taken so that the sum is \[78\] ? Explain the double answer.

From the question it is given that,

First term a = \[25\]

Common difference d = \[22\text{ }\text{ }25\text{ }=\text{ }\text{ }3\]

Sum = \[116\]

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

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

116\text{ }=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right)  \\

\end{array}\]

By cross multiplication,

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

232\text{ }=\text{ }n\text{ }\left( \left( 2\text{ }\times \text{ }25 \right)\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\text{ }\left( -3 \right) \right)  \\

232\text{ }=\text{ }n\text{ }\left( 50\text{ }\text{ }3n\text{ }+\text{ }3 \right)  \\

232\text{ }=\text{ }n\text{ }\left( 53\text{ }\text{ }3n \right)  \\

232\text{ }=\text{ }53n\text{ }\text{ }3{{n}^{2}}  \\

3{{n}^{2}}~\text{ }53n\text{ }+\text{ }232\text{ }=\text{ }0  \\

\end{array}\]

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

3{{n}^{2}}~\text{ }24n\text{ }\text{ }29n\text{ }+\text{ }232\text{ }=\text{ }0  \\

3n\text{ }\left( n\text{ }\text{ }8 \right)\text{ }\text{ }29\text{ }\left( n\text{ }\text{ }8 \right)\text{ }=\text{ }0  \\

\left( n\text{ }\text{ }8 \right)\text{ }\left( 3n\text{ }\text{ }29 \right)\text{ }=\text{ }0  \\

If\text{ }n\text{ }\text{ }8\text{ }=\text{ }0  \\

n\text{ }=\text{ }8  \\

\end{array}\]

Or

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

~3n\text{ }\text{ }29\text{ }=\text{ }0  \\

3n\text{ }=\text{ }29  \\

n\text{ }=\text{ }29/3  \\

\end{array}\]

not possible to take fraction,

So, n = \[8\]

Then, T = a + (n – 1)d

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

   =\text{ }25\text{ }+\left( 8\text{ }\text{ }1 \right)\text{ }\left( -3 \right)  \\

   =\text{ }25\text{ }+\text{ }7\text{ }\left( -3 \right)  \\

   =\text{ }25\text{ }\text{ }21  \\

   =\text{ }4  \\

\end{array}\]
(ii) How many terms of the A.P. \[\mathbf{24},\text{ }\mathbf{21},\text{ }\mathbf{18},\text{ }\ldots \] must be taken so that the sum is \[78\] ? Explain the double answer.

Solution:-

From the question it is given that,

First term a = \[24\]

Common difference d = \[21\text{ }\text{ }24\text{ }=\text{ }\text{ }3\]

Sum = \[78\]

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

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

78\text{ }=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right)  \\

\end{array}\]

By cross multiplication,

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

156\text{ }=\text{ }n\text{ }\left( \left( 2\text{ }\times \text{ }24 \right)\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\text{ }\left( -3 \right) \right)  \\

156\text{ }=\text{ }n\text{ }\left( 48\text{ }\text{ }3n\text{ }+\text{ }3 \right)  \\

156\text{ }=\text{ }n\text{ }\left( 51\text{ }\text{ }3n \right)  \\

156\text{ }=\text{ }51n\text{ }\text{ }3{{n}^{2}}  \\

\end{array}\]

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

3{{n}^{2}}~\text{ }51n\text{ }+\text{ }156\text{ }=\text{ }0  \\

3{{n}^{2}}~\text{ }12n\text{ }\text{ }39n\text{ }+\text{ }156\text{ }=\text{ }0  \\

3n\text{ }\left( n\text{ }\text{ }4 \right)\text{ }\text{ }39\text{ }\left( n\text{ }\text{ }4 \right)\text{ }=\text{ }0  \\

\left( n\text{ }\text{ }4 \right)\text{ }\left( 3n\text{ }\text{ }39 \right)\text{ }=\text{ }0  \\

\end{array}\]

If \[n\text{ }\text{ }4\text{ }=\text{ }0\]

\[n\text{ }=\text{ }4\]

or

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

3n\text{ }\text{ }39\text{ }=\text{ }0  \\

3n\text{ }=\text{ }39  \\

n\text{ }=\text{ }39/3  \\

n\text{ }=\text{ }13  \\

\end{array}\]

now we have to consider both values

So, n = \[4\]

Then,

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

T\text{ }=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d  \\

=\text{ }24\text{ }+\left( 4\text{ }\text{ }1 \right)\text{ }\left( -3 \right)  \\

=\text{ }24\text{ }+\text{ }3\text{ }\left( -3 \right)  \\

=\text{ }24\text{ }\text{ }9  \\

=\text{ }15  \\

n\text{ }=\text{ }13  \\

\end{array}\]

Then, T = a + (n – 1)d

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

=\text{ }24\text{ }+\left( 13\text{ }\text{ }1 \right)\text{ }\left( -3 \right)  \\

=\text{ }24\text{ }+\text{ }12\text{ }\left( -3 \right)  \\

=\text{ }24\text{ }\text{ }36  \\

=\text{ }-12  \\

\end{array}\]

So, \[\left( 12\text{ }+\text{ }9\text{ }+\text{ }6\text{ }+\text{ }3\text{ }+\text{ }0\text{ }+\text{ }\left( -3 \right)+\text{ }\left( -6 \right)\text{ }+\text{ }\left( -9 \right)\text{ }+\text{ }\left( -12 \right) \right)\text{ }=\text{ }0\]

Hence, the sum of \[{{5}^{th}}\] term to \[{{13}^{th}}\] term = \[0\]

i

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