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Question

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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Solution

Given: 10th term=46
a+9d=46 .........(1)
Sum of 5th term+7th term=52
a+4d+a+6d=52
2a+10d=52
a+5d=26 .........(2)
Eqn(2)Eqn(1) we get
a+5da9d=2646
4d=20
d=204=5
substituting d=5 in (1) we get
a+9×5=46
a=4645=1
a=1,d=5
The A.P is a,a+d,a+2d,a+3d,..,a+(n1)d
1,1+5,1+2×5,1+3×5,...,1+(n1)5
or 1,6,11,16,...,5n4

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