In an A. P. the 10th term is 46, sum of the 5th and 7th term is 52.Find the A. P.
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Let a and d be first term and common difference of ap
10th term=46
a+9d=46------------------------eq(1)
5th +7th term=52
a+4d+a+6d=52
2a+10d=52
a+5d=26--------------------------eq(2)
Eq(1)-eq(2)
4d=20
d=5
Put d=5 in eq(1)
a+45=46
a=1
Hence reqd ap=1,6,11,16,21,...............
10th term=46
a+9d=46------------------------eq(1)
5th +7th term=52
a+4d+a+6d=52
2a+10d=52
a+5d=26--------------------------eq(2)
Eq(1)-eq(2)
4d=20
d=5
Put d=5 in eq(1)
a+45=46
a=1
Hence reqd ap=1,6,11,16,21,...............
Answered by
5
Solution:-
Given : Sum of the 5th and 7th term is 52.
∴ a + 4d + a + 6d = 52
2a + 10d = 52 ...............(1)
Also given : 10th term is 46
∴ a + 9d = 46
Multiplying it by 2, we get
2a + 18d = 92 ................(2)
Now, subtracting equation (1) from equation (2), we get.
2a + 18d = 92
2a + 10d = 52
- - -
_______________
8d = 40
_______________
8d = 40
d = 40/8
d = 5
Substituting the value of d in equation (1),we get.
2a + 10d = 52
2a + 10*5 + 52
2a = 52 - 50
2a = 2
a = 2/2
a = 1
Therefore, the required AP is 1, 6, 11, 16, 21, 26, 31....
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