Question

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

Answer 1

Answer:

10th term/30th term = 1/3.... (given)

a+9d/a+29d = 1/3

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

3a+27d = a+29d

3a-a+27d-29d = 0

2a-2d = 0

a-d = 0....(1)

sum of 6 terms = n/2[2a+(n-1)d]

                    42 = 6/2[2a+(6-1)d]

                    42 = 3[2a+5d]

                    42 = 6a+15d

                     2a+5d = 14....(2)

Subtracting (1) from (2)

  2a+5d=14

(-)1a-1d   =0

-     +        -

a+6d=14

a=14-6d...(3)

Substituting (3) in (1)

14-6d-d=0

14-7d=0

14=7d

d=14/7

d=2

Substituting value of 'd' in (3)

a=14-6(2)

a=14-12

a=2

$a_{1}$=2

$a_{13}$= a+12d

    = 2+12(2)

    = 2+24

$a_{13}$= 26

Answer 2

Given Question:-

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

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•Concept :-

•Here, the concept of nth term of AP has been used for finding required solution.

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•Solution :-

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Let

•a=first term of Arithematic progression

•d=common difference of the Arithematic progression

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Now ,According to question:-

$\bull \bf \: a _{1} + a _{2} + a _{3} + a _{4} + a _{5} = 42$

Now using below formula

$\bull \bf \: a _{n} = a + (n - 1)d$

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We get,

$\implies \bf a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d) = 42$

$\implies \bf \: \: 6a + 15d = 42$

Dividing by 3 on both side,we get

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

And,

$\bf \longrightarrow \: \: \: \: \dfrac{a _{10} }{a _{30} } = \dfrac{1}{3}$

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•By cross-multiplication we get:

$\implies \bf \: \: \: 3a _{10} = a _{30}$

Again using below formula

$\bull \bf \: a _{n} = a + (n - 1)d$

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We get:

$\bf \implies3(a + 9d) = (a + 29d)$

$\implies \bf \: \: 3a + 27d = a + 29d$

$\implies \bf 3a -a = 29d - 27d$

$\implies \bf2a = 2d$

$\implies \bf \: a = d$

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Now putting a=d in equation (1) we get

$\bf \implies2a + 5a = 14$

$\implies \bf a = 2$

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Now a=d=2

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Hence, the first term of Arithematic progression is 2

Now its thirteenth term

$\bf \: a _{13} = a + (13 - 1)d$

$\: \: \: \: \: \bf = a + 12d$

$\: \: \: \: \: \bf = 2 + 12(2)$

$\: \: \: \: \: \bf = 26$

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Hence, the thirteenth term of Arithematic progression is 26