Math, asked by amankoli200, 1 year ago

the sum of first 6 term of an ap is 42. the ratio its 10th term to 38th term is 1:3 .calculate thr first and 13th term of the AP​

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Answered by Yashraj1234
3

Answer:d receive 10 points!

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The sum of first six terms of an AP is 42. The ratio of 10th term to its 30th term is 1:3.calculate the fist term and 13th term of A . SORRY it's maths question

Report by Nirali19 26.12.2016

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THE BRAINLIEST ANSWER!

TheRuhanikaDhawan

TheRuhanikaDhawan Chemistry Lavoiser

S6 =42


a + 9d 1

------------ = -------

a + 29d 3


cross multiply we get


3a + 27d = a +29 d


2a - 2d = 0 ------------------ (1)

its given that

sum of first six terms of an AP is 42


therefore


S6 = n/2 ( 2a + (n-1) d)


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


42 = 3 (2a + 5d )


14 = 2a +5d


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


solve eq 1 and 2


2a - 2d = 0

2a +5d = 14


we get

d= 2

a = 2

---------------

13th term of AP

= a + (n-1)d


2+ (13-1) 2


= 2+ 24


=26

-----------------------------------------tep-by-step explanation:



Yashraj1234: Ruhanika is my cousin that is why i wrote her name
amankoli200: no no 38 not 30
Answered by nostacovspitar
1

Imo, it is the thirtieth term (in the question) and not the thirty-eigth term)


Given that

S6 = 6/2 (a + a6)

S6 = 3(a + a + 5d)................(a6 = a +5d)

42 = 3(2a + 5d)

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


a10/a30 = 1/3

a+9d/a + 29d = 1/3.........{an = a + (n-1)d}

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

3a + 27d = a + 29d

3a - a = 29d - 27d

2a = 2d

a = d


Substituting in (i)


14 = 2a + 5d = 2a + 5a = 7a

a = 2

d = 2

a13 = a + 12d = 2 + 12(2) = 2 + 24

a13 = 26


Hence, the first and the thirteenth terms of the AP are 2 and 26 respectively



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