The ratio of the sums of $m$ and $n$ terms of an $A.P.$ is $m^{2}: n^{2} .$ Show that the ratio of $m^{ th }$ and $n^{ th }$ term is $(2 m-1):(2 n-1)$
Let $a$ and $b$ be the first term and the common difference of the $A.P.$ respectively. According to the given condition,
$\frac{{{\rm{ Sum}}\,\,{\rm{of }}\,\,m\,\,{\rm{ terms }}}}{{{\rm{ Sum }}\,\,{\rm{of}}\,{\rm{ }}n{\rm{ }}\,\,{\rm{terms }}}} = \frac{{{m^2}}}{{{n^2}}}$
$\Rightarrow \frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^{2}}{n^{2}}$
$\Rightarrow \frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}$ ........$(1)$
Putting $m=2 m-1$ and $n=2 n-1,$ we obtain
$\frac{2 a+(2 m-2) d}{2 a+(2 n-2) d}=\frac{2 m-1}{2 n-1}$
$\Rightarrow \frac{a+(m-1) d}{a+(n-1) d}=\frac{2 m-1}{2 n-1}$ ..........$(2)$
$\frac{{{m^{th}}\,\,{\rm{ term}}\,\,{\rm{ of}}\,\,{\rm{ A}}{\rm{.P}}{\rm{. }}}}{{{n^{{\rm{th }}}}\,\,{\rm{ term }}\,\,{\rm{of}}\,\,{\rm{ A}}{\rm{.P}}{\rm{. }}}} = \frac{{a + (m - 1)d}}{{a + (n - 1)d}}$
From $(2)$ and $(3),$ we obtain
$\frac{m^{H h} \text { termof A.P. }}{n^{t h} \text { termof A.P. }}=\frac{2 m-1}{2 n-1}$
Thus, the given result is proved.
The ratio of the sums of first $n$ even numbers and $n$ odd numbers will be
If ${a_1},\;{a_2},\,{a_3},......{a_{24}}$ are in arithmetic progression and ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$, then ${a_1} + {a_2} + {a_3} + ........ + {a_{23}} + {a_{24}} = $
The sum of the first four terms of an $A.P.$ is $56 .$ The sum of the last four terms is $112.$ If its first term is $11,$ then find the number of terms.
The sum of the common terms of the following three arithmetic progressions.
$3,7,11,15,...................,399$
$2,5,8,11,............,359$ and
$2,7,12,17,...........,197$, is equal to $................$.
If $\alpha ,\;\beta ,\;\gamma $ are the geometric means between $ca,\;ab;\;ab,\;bc;\;bc,\;ca$ respectively where $a,\;b,\;c$ are in A.P., then ${\alpha ^2},\;{\beta ^2},\;{\gamma ^2}$ are in