The ratio of the sums of first m and first n terms of an A. P. is \[{{m}^{2}}:{{n}^{2}}\]. |
Show that the ratio of its mth and nth terms is \[2(m-1):(2n-1)\]. |
Answer:
Let a be first term and d is common difference. Then, \[\frac{{{S}_{m}}}{{{S}_{n}}}=\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[n[2a+(m-1)d]=m[2a+(n-1)d]\] \[\Rightarrow \] \[2an+nd(m-1)=2am+md(n-1)\] \[\Rightarrow \] \[2a(n-m)=[m(n-1)-n(m-1)d\] \[=(mn-m-mn+n)d\] \[2a(n-n)=(n-m)d\] \[2a=d\] Now, \[\frac{{{T}_{m}}}{{{T}_{n}}}=\frac{a+(m-1)d}{a+(n-1)d}\] \[=\frac{a+(m-1)2a}{a+(n-1)2a}=\frac{2m-1}{2n-1}\] Hence Proved.
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