Question

the sum of first six terms of an AP is 42. The ratio of 10th term to its 38th term is 1:3.calculate the fist term and 13th term of the AP .​

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Answer 1

Answer: First-term = 14/3, 13th term = 238/15

Step-by-step explanation:

The sum of the first 6 terms = 42

$Sum = (n/2)(2a+(n-1)d)\\Here, n = 6, sum = 42\\42 = (6/2)(2a+5d)\\2a+5d = 14...........(1)$

Ratio of 10th and 38th term = 1:3

$(a+9d)/(a+37d) = 1/3\\a = 5d.........(2)$

From (1) and (2);

a = 14/3 = First term

d = 14/15

So, 13th term = a+12d = 238/15

(I hope it will be marked as Brainliest)

Answer 2

$Let \: a \: and \: d \: are \: first \:term \:and$

$common \: difference \: of \: an \: A.P$

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

$Here , n = 6 \:and\: S_{6} = 42$

$\implies \frac{6}{2}[ 2a +5d ] = 42$

$\implies 3(2a+5d) = 42$

$\implies 2a + 5d = \frac{42}{3}$

$\implies 2a + 5d = 14 \: --(1)$

/* We know that, */

$\boxed{ \blue{n^{th} \:term (a_{n}) = a + (n-1)d }}$

$Ratio \: of \: 10^{th} \: term \: and$

$38^{th} \:term\: = \frac{1}{3}$

$\implies \frac{a+9d}{a+37d} = \frac{1}{3}$

$\implies 3(a+9d) = a+37d$

$\implies 3a+ 27d= a+37d$

$\implies 3a - a = 37d - 27d$

$\implies 2a = 10d \: --(2)$

$Put \: 2a = 10d \: in \: equation \: (1), we \:get$

$\implies 10d + 5d = 14$

$\implies 15d = 14$

$\implies d = \frac{14}{15} \: --(3)$

$Put \: d = \frac{14}{15}\: in\: equation \:(2) ,$

$we \:get$

$2a = 10 \times \frac{14}{15}$

$\implies a = 2 \times \frac{7}{3}$

$\implies a = \frac{14}{3} \: --(4)$

$13^{th} \:term (a_{13} ) = a + 12d$

$= \frac{14}{3} + 12 \times \frac{14}{15}$

$= \frac{14}{3} + 4 \times \frac{14}{5}$

$= \frac{14}{3} + \frac{56}{5}$

$= \frac{70+168}{15}$

$= \frac{238}{15}$

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