The sum of the first six terms of an arithmetic series is 255 and that of the first nine terms is 315. Find the first four terms of the series.
Ans.=55+50+45+40+........
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Answered by
6
ans. is 0 . Sum of four terms = 190(55+50+45+40+0) first term of AP will be 55 and common diff. will be( –5).
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Answered by
31
Simple.
A.P : a, a+d, a+2d, a+3d, ..... a+(n-1)d
Sum of first n terms: Sn = [ 2 a + (n-1) d ] * n/2
Given S₆ = [ 2 a + 5 d ] * 6/2 = 255
=> 2 a + 5 d = 85 --------(1)
Given S₉ = [ 2 a + 8 d ] * 9/2 = 315
=> 2 a + 8 d = 630/9 = 70 --- (2)
(1) - (2) => - 3 d = 15
d = - 5
From eq (1), 2 a = 85 - 5 d = 85 + 25 = 110
a = 55
A.P: 55 + (55-5) + (55-10) + (55-15) + (55-20) + (55-25) + ...
55 + 50 + 45 + 40 + 35 + 30 + ....
Sum of first FOUR terms: 55+50+45+40 = 190.
A.P : a, a+d, a+2d, a+3d, ..... a+(n-1)d
Sum of first n terms: Sn = [ 2 a + (n-1) d ] * n/2
Given S₆ = [ 2 a + 5 d ] * 6/2 = 255
=> 2 a + 5 d = 85 --------(1)
Given S₉ = [ 2 a + 8 d ] * 9/2 = 315
=> 2 a + 8 d = 630/9 = 70 --- (2)
(1) - (2) => - 3 d = 15
d = - 5
From eq (1), 2 a = 85 - 5 d = 85 + 25 = 110
a = 55
A.P: 55 + (55-5) + (55-10) + (55-15) + (55-20) + (55-25) + ...
55 + 50 + 45 + 40 + 35 + 30 + ....
Sum of first FOUR terms: 55+50+45+40 = 190.
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