The sum of first $6$ terms of an $A P$ is $42 .$ The ratio of its $11 t h$ term to its $33 t h$ term is $1 : 3 .$ Calculate the first term of an $A . P .$

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Solution

Formula,

$a_{n} = a + (n - 1) d$

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Given,

$\frac{a_{11}}{a_{33}} = \frac{1}{3}$

$\frac{a + 10 d}{a + 32 d} = \frac{1}{3}$

$3 a + 30 d = a + 32 d$

$2 a = 2 d$

$\Rightarrow a = d$

$S_{6} = \frac{6}{2} [2 a + (6 - 1) d] = 42$

$2 a + 5 d = 14$

$7 a = 14$

$∴ a = 2$

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The sum of first 6 terms of an AP is 42. The ratio of its 10th term to its 30th term is 1:3 . Calculate the first and the thirteenth term of the AP.

The sum of the first six terms of an AP is 42.the ratio of its 10th term to its 13th term is 1:3 calculate the first and the thirteenth term of an AP

A P

42

10^{t h}

30^{t h}

1 : 3

13^{t h} t e r m

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42

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30^{t h}

1 : 3

30^{t h}

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