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Question 7
The sum of the 5th and the 7th terms of an AP is 52and the 10th term is 46. Find the AP.


Solution

Let the first and common difference of AP are a and d, respectively.
According to the question,
a5+a7=52 and a10=46
     a+(51)d+a+(71)d=52      [an=a+(n1)d]
And           a + (10 - 1)d = 46
    a + 4d + a + 6d = 52
And    a+ 9d = 46
    2a + 10d = 52
And   a + 9d = 46
   a + 5d = 26        . . . . .  .(i)
        a + 9d = 46            . . . . . .  .(ii)
On subtracting eq.(i) from eq. (ii), we get;
4d = 20 d = 5
From eq. (i),               a = 26 - 5(5) = 1
So, required AP is a, a + d, a + 2d, a + 3d, . . . . i.e, 1, 1 + 5, 1 + 2(5) , 1 + 3(5), . . . . i.e, 1, 6, 11, 16,....

Mathematics
NCERT Exemplar
Standard X

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