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Home » The sum of first six terms of an arithmetic progression is     . The ratio of the     term to the     term is     . Calculate the first and the thirteenth term.
The sum of first six terms of an arithmetic progression is

    \[42\]

. The ratio of the

    \[{{10}^{th}}\]

term to the

    \[\mathbf{3}{{\mathbf{0}}^{\mathbf{th}}}\]

term is

    \[1:3\]

. Calculate the first and the thirteenth term.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
The sum of first six terms of an arithmetic progression is

    \[42\]

. The ratio of the

    \[{{10}^{th}}\]

term to the

    \[\mathbf{3}{{\mathbf{0}}^{\mathbf{th}}}\]

term is

    \[1:3\]

. Calculate the first and the thirteenth term.

Given,

    \[{{S}_{6}}~=\text{ }42\text{ }and\text{ }{{T}_{10}}/{{T}_{30}}~=\text{ }1/3\]

We know that,

    \[\begin{array}{*{35}{l}} {{S}_{n}}~=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right) \\ {{T}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ \end{array}\]

So, we have

    \[\begin{array}{*{35}{l}} {{S}_{6}}~=\text{ }\left( 6/2 \right)\text{ }\times \text{ }\left[ 2a\text{ }+\text{ }\left( 6\text{ }\text{ }1 \right)d \right]\text{ }=\text{ }42 \\ 3\text{ }\times \text{ }\left( 2a\text{ }+\text{ }5d \right)\text{ }=\text{ }42 \\ \end{array}\]

    \[2a\text{ }+\text{ }5d\text{ }=\text{ }14\]

… (i) [Dividing by

    \[3\]

]

And,

    \[\begin{array}{*{35}{l}} {{T}_{10}}/{{T}_{30}}~=\text{ }\left[ a\text{ }+\text{ }\left( 10\text{ }\text{ }1 \right)d \right]/\text{ }\left[ a\text{ }+\text{ }\left( 30\text{ }\text{ }1 \right)d \right]\text{ }=\text{ }1/3 \\ \left( a\text{ }+\text{ }9d \right)/\text{ }\left( a\text{ }+\text{ }29d \right)\text{ }=\text{ }1/3 \\ \end{array}\]

On cross-multiplication, we have

    \[\begin{array}{*{35}{l}} 3\left( a\text{ }+\text{ }9d \right)\text{ }=\text{ }a\text{ }+\text{ }29d \\ 3a\text{ }+\text{ }27d\text{ }=\text{ }a\text{ }+\text{ }29d \\ 3a\text{ }\text{ }a\text{ }+\text{ }27d\text{ }\text{ }29d\text{ }=\text{ }0 \\ 2a\text{ }\text{ }2d\text{ }=\text{ }0 \\ a\text{ }\text{ }d\text{ }=\text{ }0 \\ a\text{ }=\text{ }d \\ \end{array}\]

Using the above the relation in (i), we get

    \[\begin{array}{*{35}{l}} 2a\text{ }+\text{ }5a\text{ }=\text{ }14 \\ 7a\text{ }=\text{ }14 \\ a\text{ }=\text{ }14/7 \\ a\text{ }=\text{ }2 \\ \end{array}\]

Hence, the first term of the A.P is

    \[2\]

.

Now, the thirteenth term is given by

    \[\begin{array}{*{35}{l}}</strong> <strong> {{T}_{13}}~=\text{ }2\text{ }+\text{ }\left( 13\text{ }\text{ }1 \right)\left( 2 \right) \\</strong> <strong> =\text{ }2\text{ }+\text{ }12\text{ }\times \text{ }2 \\</strong> <strong> =\text{ }2\text{ }+\text{ }24 \\</strong> <strong> =\text{ }26 \\</strong> <strong>\end{array}\]

i

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