If sum of 3rd and 8th terms of an ap us 7 and sum of 7th and 14th terms is - 3 then find the 10th term
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FORMULA
an = a + (n - 1)d
Where an = (last term)
a = first term
d = common difference
n = number of terms
GIVEN
sum of 3rd and 8th terms of an A. P is 7
a3 + a8 = 7
(a + 2d) + (a + 7d) = 7
2a + 9d = 7........... (1)
sum of 7th and 14th terms of an A. P
is - 3
a7 + a14 = - 3
(a + 6d) + (a + 13d) = - 3
2a + 19d = - 3............. (2)
SUBTRACT (1) from (2)
(2a + 19d = - 3.) - (2a + 9d = 7)
(2a - 2a) + (19d - 9d) = (- 3 - 7)
10d = - 10
d = - 10 / 10
d = - 1
SUBSITUTE d value in any (1) or (2)., we will Subsitute in(2)
2a + [19 x (- 1)] = - 3
2a - 19 = - 3
2a = - 3 + 19
2a = 16
a = 16 / 2
a = 8
BY FORMULA
a10 = a + 9d
a10 = 8 + [9 x (- 1)]
a10 = 8 - 9
a10 = - 1
Therefore, 10th term 'a10' is - 1.
an = a + (n - 1)d
Where an = (last term)
a = first term
d = common difference
n = number of terms
GIVEN
sum of 3rd and 8th terms of an A. P is 7
a3 + a8 = 7
(a + 2d) + (a + 7d) = 7
2a + 9d = 7........... (1)
sum of 7th and 14th terms of an A. P
is - 3
a7 + a14 = - 3
(a + 6d) + (a + 13d) = - 3
2a + 19d = - 3............. (2)
SUBTRACT (1) from (2)
(2a + 19d = - 3.) - (2a + 9d = 7)
(2a - 2a) + (19d - 9d) = (- 3 - 7)
10d = - 10
d = - 10 / 10
d = - 1
SUBSITUTE d value in any (1) or (2)., we will Subsitute in(2)
2a + [19 x (- 1)] = - 3
2a - 19 = - 3
2a = - 3 + 19
2a = 16
a = 16 / 2
a = 8
BY FORMULA
a10 = a + 9d
a10 = 8 + [9 x (- 1)]
a10 = 8 - 9
a10 = - 1
Therefore, 10th term 'a10' is - 1.
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