plz tell me............. This sum of the first N terms of an ap is 153 and the common difference is 2.If the first term is an integer ,then what is number of possible values of n.
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The sum of n terms of an AP is a well known formula. Using that, we can conveniently derive the quadratic equation we need to investigate.
Sn=n2[2a+(n−1)d]Sn=n2[2a+(n−1)d]
Here, 153=n2[2a+(n−1)2]153=n2[2a+(n−1)2]
or, 153=an+(n−1)n153=an+(n−1)n
or, n2+(a−1)n−153=0n2+(a−1)n−153=0
From the quadratic formula, we can write : n=1−a±(a−1)2+612−−−−−−−−−−−√2n=1−a±(a−1)2+6122
or n=(1−a)2±[a−12]2+153−−−−−−−−−−−−−⎷n=(1−a)2±[a−12]2+153
This is where the question gets tricky, as we have to predict when will the square root give us an integer and also check that nn remains positive. The latter half can be done, once we narrow down the possible values of aa.
Let a−12=pa−12=p , some integer. Then we need values of pp and qq (another integer) such that p2+153=q2p2+153=q2
or 153=q2−p2153=q2−p2
Now, here we need to know the following approach. A number can be expressed as difference of 22 squares, if it can be expressed as a product of 22 odd factors, or that of 22 even factors. This follows from the following property.
if x=abx=ab,
then x=[a+b2]2−[a−b2]2x=[a+b2]2−[a−b2]2
In our case, we have to express 153153 as product of either 22 odds, or 22 evens. Clearly, there are 22possibilities : 3×513×51 and 9×179×17
From here, we can get the value of q=27q=27 and 1313, which in turn gives us the values of aa as 4949 and 99 respectively. These values can now get us the value of nn. It can be calculated as nn = 33 and 99, considering only the positive values, of course.
So, the total number of possible values that nncan take is 2
Sn=n2[2a+(n−1)d]Sn=n2[2a+(n−1)d]
Here, 153=n2[2a+(n−1)2]153=n2[2a+(n−1)2]
or, 153=an+(n−1)n153=an+(n−1)n
or, n2+(a−1)n−153=0n2+(a−1)n−153=0
From the quadratic formula, we can write : n=1−a±(a−1)2+612−−−−−−−−−−−√2n=1−a±(a−1)2+6122
or n=(1−a)2±[a−12]2+153−−−−−−−−−−−−−⎷n=(1−a)2±[a−12]2+153
This is where the question gets tricky, as we have to predict when will the square root give us an integer and also check that nn remains positive. The latter half can be done, once we narrow down the possible values of aa.
Let a−12=pa−12=p , some integer. Then we need values of pp and qq (another integer) such that p2+153=q2p2+153=q2
or 153=q2−p2153=q2−p2
Now, here we need to know the following approach. A number can be expressed as difference of 22 squares, if it can be expressed as a product of 22 odd factors, or that of 22 even factors. This follows from the following property.
if x=abx=ab,
then x=[a+b2]2−[a−b2]2x=[a+b2]2−[a−b2]2
In our case, we have to express 153153 as product of either 22 odds, or 22 evens. Clearly, there are 22possibilities : 3×513×51 and 9×179×17
From here, we can get the value of q=27q=27 and 1313, which in turn gives us the values of aa as 4949 and 99 respectively. These values can now get us the value of nn. It can be calculated as nn = 33 and 99, considering only the positive values, of course.
So, the total number of possible values that nncan take is 2
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