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Home » If the sum of first     terms of an A.P. is     and that of the first     terms is     , find the sum of first 10 terms.
If the sum of first

    \[6\]

terms of an A.P. is

    \[36\]

and that of the first

    \[16\]

terms is

    \[256\]

, find the sum of first 10 terms.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
If the sum of first

    \[6\]

terms of an A.P. is

    \[36\]

and that of the first

    \[16\]

terms is

    \[256\]

, find the sum of first 10 terms.

From the question it is given that,

    \[\begin{array}{*{35}{l}} {{S}_{6}}~=\text{ }36 \\ {{S}_{16}}~=\text{ }256 \\ \end{array}\]

We know that,

    \[\begin{array}{*{35}{l}} {{S}_{n}}~=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right) \\ {{S}_{6}}~=\text{ }\left( 6/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( 6\text{ }\text{ }1 \right)d \right)\text{ }=\text{ }36 \\ 3\text{ }\left( 2a\text{ }+\text{ }5d \right)\text{ }=\text{ }36 \\ \end{array}\]

Divide both the side by

    \[3\]

,

    \[2a\text{ }+\text{ }5d\text{ }=\text{ }12\]

… [equation (i)]

Now,

    \[\begin{array}{*{35}{l}} ~{{S}_{16}}~=\text{ }\left( 16/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( 16\text{ }\text{ }1 \right)d \right)\text{ }=\text{ }256 \\ 8\text{ }\left( 2a\text{ }+\text{ }15d \right)\text{ }=\text{ }256 \\ \end{array}\]

Divide both the side by

    \[8\]

,

    \[2a\text{ }+\text{ }15d\text{ }=\text{ }32\]

… [equation (ii)]

Then, subtract equation (ii) from equation (i) we get,

    \[\begin{array}{*{35}{l}} \left( 2a\text{ }+\text{ }5d \right)\text{ }\text{ }\left( 2a\text{ }+\text{ }15d \right)\text{ }=\text{ }12\text{ }\text{ }32 \\ 2a\text{ }+\text{ }5d\text{ }\text{ }2a\text{ }\text{ }15d\text{ }=\text{ }-20 \\ -10d\text{ }=\text{ }-20 \\ d\text{ }=\text{ }-20/-10 \\ d\text{ }=\text{ }2 \\ \end{array}\]

substitute the value of d in equation (i) to find a,

    \[\begin{array}{*{35}{l}} 2a\text{ }+\text{ }5d\text{ }=\text{ }12 \\ 2a\text{ }+\text{ }5\left( 2 \right)\text{ }=\text{ }12 \\ 2a\text{ }+\text{ }10\text{ }=\text{ }12 \\ 2a\text{ }=\text{ }12\text{ }\text{ }10 \\ 2a\text{ }=\text{ }2 \\ a\text{ }=\text{ }2/2 \\ a\text{ }=\text{ }1 \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}} {{S}_{10}}~=\text{ }\left( n/2 \right)\text{ }\left( 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right) \\ =\text{ }\left( 10/2 \right)\text{ }\left( \left( 2\text{ }\times \text{ }1 \right)\text{ }+\text{ }\left( 10\text{ }\text{ }1 \right)2 \right) \\ =\text{ }5\text{ }\left( 2\text{ }+\text{ }18 \right) \\ =\text{ }5\text{ }\left( 20 \right) \\ =\text{ }100 \\ \end{array}\]

Therefore, the sum of first

    \[10\]

terms is

    \[100\]

.

i

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