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Home » If the third term of an A.P. is     and the ratio of its 6th term to the 10th term is     , thenfind the sum of first     terms of this A.P.
If the third term of an A.P. is

    \[5\]

and the ratio of its 6th term to the 10th term is

    \[7:13\]

, thenfind the sum of first

    \[20\]

terms of this A.P.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
If the third term of an A.P. is

    \[5\]

and the ratio of its 6th term to the 10th term is

    \[7:13\]

, thenfind the sum of first

    \[20\]

terms of this A.P.

From the question it is given that,

The third term of an A.P. a3 =

    \[5\]

The ratio of its 6th term to the 10th term

    \[~{{a}_{6}}~:\text{ }{{a}_{10}}~=\text{ }7\text{ }:\text{ }13\]

We know that, an = a + (n – 1)d

    \[{{a}_{3}}~=\text{ }a\text{ }+\text{ }\left( 3\text{ }\text{ }1 \right)d\text{ }=\text{ }5\]

= a + 2d = 5 … [equation (i)]

Then,

    \[{{a}_{6}}/{{a}_{10}}~=\text{ }7/13\]

    \[\left( a\text{ }+\text{ }5d \right)/\left( a\text{ }+\text{ }9d \right)\text{ }=\text{ }7/13\]

By cross multiplication we get,

    \[\begin{array}{*{35}{l}} 13\left( a\text{ }+\text{ }5d \right)\text{ }=\text{ }7\left( a\text{ }+\text{ }9d \right) \\ 13a\text{ }+\text{ }75d\text{ }=\text{ }7a\text{ }+\text{ }63d \\ 13a\text{ }\text{ }7a\text{ }+\text{ }65d\text{ }\text{ }63d\text{ }=\text{ }0 \\ 6a\text{ }+\text{ }2d\text{ }=\text{ }0 \\ \end{array}\]

Divide by

    \[2\]

on both side we get,

    \[3a\text{ }+\text{ }d\text{ }=\text{ }0\]

    \[d=-3a\]

… [equation (ii)]

Substitute the value of d in equation (i),

    \[\begin{array}{*{35}{l}} a\text{ }+\text{ }2\left( -3a \right)\text{ }=\text{ }5 \\ a\text{ }\text{ }6a\text{ }=\text{ }5 \\ -5a\text{ }=\text{ }5 \\ a\text{ }=\text{ }-5/5 \\ a\text{ }=\text{ }-1 \\ \end{array}\]

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

    \[\begin{array}{*{35}{l}} d\text{ }=\text{ }-3\left( -1 \right) \\ d\text{ }=\text{ }3 \\ \end{array}\]

Then, sum of first

    \[20\]

terms,

    \[\begin{array}{*{35}{l}}</strong> <strong> =\text{ }\left( n/2 \right)\text{ }\left[ 2a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \right] \\</strong> <strong> =\text{ }\left( 20/2 \right)\left[ \left( 2\text{ }\times \text{ }\left( -3 \right) \right)\text{ }+\text{ }\left( 2-\text{ }\text{ }1 \right)3 \right] \\</strong> <strong> =\text{ }10\left[ -2\text{ }+\text{ }57 \right] \\</strong> <strong> =\text{ }10\text{ }\times \text{ }55 \\</strong> <strong> =\text{ }550 \\</strong> <strong>\end{array}\]

i

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