Since\( 2 = a_n –a_{n-1}\)
diff=2
Hence\( a_1=1, a_2=3,a_3=5\)
For n terms,\( S_n=\frac{n*(2*1+(n-1)*2)}{2}\)
=\(n^2\)
Similarly,\(S_m=[m]m^2\)
\( S_m-S_n= m^2-n^2=65\)
\((m-n)(m+n)=65\)
There are two possible sets of 65={1,65},{5,13}
Therefore,
m-n=1
m+n=65
and
m-n=5
m+n= 13
Hence Answer is D.. 2Pairs
diff=2
Hence\( a_1=1, a_2=3,a_3=5\)
For n terms,\( S_n=\frac{n*(2*1+(n-1)*2)}{2}\)
=\(n^2\)
Similarly,\(S_m=[m]m^2\)
\( S_m-S_n= m^2-n^2=65\)
\((m-n)(m+n)=65\)
There are two possible sets of 65={1,65},{5,13}
Therefore,
m-n=1
m+n=65
and
m-n=5
m+n= 13
Hence Answer is D.. 2Pairs
Bunuel wrote:
The sequence\(a_1\) ,\(a_2\) , ...\(a_n\) , ... is such that\( 2 = a_n –a_{n-1}\) for all positive integers\( n ≥2\) . If\(S_n\) denotes the sum of the first n terms of the sequence and\( a_3 =5\) , how many pairs (m, n) exist such than
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