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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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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TheRuhanikaDhawan
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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:
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