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Determine the A.P. whose third term is

    \[16\]

and the

    \[{{7}^{th}}\]

term exceeds the

    \[{{5}^{th}}\]

term by

    \[12\]

.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
Determine the A.P. whose third term is

    \[16\]

and the

    \[{{7}^{th}}\]

term exceeds the

    \[{{5}^{th}}\]

term by

    \[12\]

.

From the question it is given that,

    \[{{T}_{3}}~=\text{ }16\]

The 7th term exceeds the 5th term by

    \[12\text{ }=\text{ }{{T}_{7}}~\text{ }{{T}_{5}}~=\text{ }12\]

We know that,

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

So,

    \[{{T}_{3}}~=\text{ }a\text{ }+\text{ }2d\text{ }=\text{ }16\]

… [equation (i)]

    \[{{T}_{7}}~\text{ }{{T}_{5~}}=\text{ }\left( a\text{ }+\text{ }6d \right)\text{ }\text{ }\left( a\text{ }+\text{ }4d \right)\text{ }=\text{ }12\]

… [equation (ii)]

    \[\begin{array}{*{35}{l}} 12\text{ }=\text{ }a\text{ }+\text{ }6d\text{ }\text{ }a\text{ }\text{ }4d \\ 12\text{ }=\text{ }2d \\ d\text{ }=\text{ }12/2 \\ d\text{ }=\text{ }6 \\ \end{array}\]

Now, substitute value of d in equation (i) we get,

Then,

    \[\begin{array}{*{35}{l}} {{T}_{3}}~=\text{ }a\text{ }+\text{ }2d \\ 16\text{ }=\text{ }a\text{ }+\text{ }2\left( 6 \right) \\ a\text{ }=\text{ }16\text{ }\text{ }12 \\ a\text{ }=\text{ }4 \\ \end{array}\]

Therefore, A.P. is

    \[4\text{ }+\text{ }6\text{ }=\text{ }10,\text{ }10\text{ }+\text{ }6\text{ }=\text{ }16,\text{ }16\text{ }+\text{ }6\text{ }=\text{ }22\]

Hence, the four term of A.P. is

    \[4,\text{ }10,\text{ }16,\text{ }22,\text{ }\ldots .\]

i

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