The sum of the 4th term and 6th terms of an arithmetic progression is 42.the sum of the 3rd and 9th term of the progression is 52. Find the first term, the common difference and the sum of the first 10th terms of the progression
Answers
Given,
4th term + 6th term = 42 ----------(1)
3rd term + 9th term = 52 ----------(2)
From equation 1,
a + 3d + a + 5d = 42. ---------(3)
From equation 2,
a + 2d + a + 8d = 52 ----------(4)
Subtracting equation 3 and 4,
2a + 10d = 52
2a + 8d = 42
(-). (-). (-)
--------------------------------
2d = 10
d = 5
Therefore, common difference d = 5.
Now, substituting the value of d=5 in equation(3)
2a + 8d = 42
2a + 8*5 = 42
2a + 40 = 42
2a = 42 - 40
a = 2/2
a = 1
Sum of the first 10 terms,
S = n/2{ a + (n - 1)d}
= 10/2{ 1 + 9 * 5}
= 5( 1+ 45)
= 5 * 46
= 230
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