Sum of first six terms of an $A P$ is $42$ . The ratio of its $10^{t h}$ term to $30^{t h}$ term is $1 : 3$ find the first and $13^{t h} t e r m$ of an $A P$ .

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Solution

Sum of the first n terms is given by

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

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

$S_{6} = 3 [2 a + 5 d]$

$\frac{T_{10}}{T_{30}} = \frac{1}{3}$

$\frac{a + 9 d}{a + 29 d} = \frac{1}{3}$

$(a + 9 d) 3 = (a + 29 d) 1$

$3 a + 27 d = a + 29 d$

$2 a = 2 d$

$a = d$ ...(1)

$S_{6} = 3 [2 a + 5 d]$

$S_{6} = 3 [2 a + 5 a]$

$42 = 3 (7 a)$

$42 = 21 a$

$∴ a = 2$

From (1)

$d = a = 2$

$T_{13} = a + (n - 1) d$

$= 2 + (13 - 1) 2$

$= 2 + 24$

$= 26$

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