The sum of three terms which are in an AP is 33, if the product of first term and third term exceeds the second term by 29 find the AP.
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Hey .. I hope my soln will satisfied you....
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shalini75:
This is the consecutive term in the AP
I m sry ..what do u mean by -d
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Let the three terms be a-d, a, a+d.
Given that sum of three terms which are in AP = 33.
a + a - d + a + d = 33
3a = 33
a = 11.
Given that the product of the first term and third term exceeds the second term.
(a - d)(a + d) = 29 + a ------- (2)
Substitute a = 11 in (2), we get
(11 - d)(11 + d) = 29 + 11
11^2 - d^2 = 40
121 - d^2 = 40
-d^2 = -81
d = 9 (or) -9.
When d = 9,
a = 11
a - d = 11 - 9 = 2
a + d = 11 + 9 = 20.
When d = -9
a = 11
a - d = 11 - (-9) = 20
a + d = 11 + (-9) = 2.
Therefore the AP is 20,11,2 (or) 2,11,20.
Hope this helps!
Given that sum of three terms which are in AP = 33.
a + a - d + a + d = 33
3a = 33
a = 11.
Given that the product of the first term and third term exceeds the second term.
(a - d)(a + d) = 29 + a ------- (2)
Substitute a = 11 in (2), we get
(11 - d)(11 + d) = 29 + 11
11^2 - d^2 = 40
121 - d^2 = 40
-d^2 = -81
d = 9 (or) -9.
When d = 9,
a = 11
a - d = 11 - 9 = 2
a + d = 11 + 9 = 20.
When d = -9
a = 11
a - d = 11 - (-9) = 20
a + d = 11 + (-9) = 2.
Therefore the AP is 20,11,2 (or) 2,11,20.
Hope this helps!
U can take a,a-d,a+2d (or) a+2d,a,a+d (or) any value its ur wish. Since it is AP.
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