The sum of first 6 terms of an A.P is 42 the ratio of its 10th term and 13th term is 1:3 find 12th term
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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
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
Anikvish842:
But here is given that the ratio between term 10 and term 13 not term 30
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0
Answer:
Step-by-step explanation:
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