Question

If \[{(m + 1)^{th}}\] term of an A.P. is twice the \[{(n + 1)^{th}}\] term, prove that \[{(3m + 1)^{th}}\] term is twice that of \[{(m + n + 1)^{th}}\] term.

Answer 1

Hint: From the given equality, we will first find the common difference in terms of first term \[a\], \[m\] and \[n\]. Then we find \[{(3m + 1)^{th}}\] term and \[{(m + n + 1)^{th}}\] term by using the value of common difference and then make a relation between the two terms of A.P

Complete step-by-step answer:
The given condition says that \[{(m + 1)^{th}}\] term of an A.P. is twice the \[{(n + 1)^{th}}\] term, so on converting this into an equation by using the formula of \[{n^{th}}\] term of A.P, it can be written as
The formula for \[{n^{th}}\] term of an A.P is \[t = a + (n - 1)d\], where \[a\] is the first term of the A.P and \[d\] is the common difference .
Thus, writing an equation as per the first condition, we get
\[
   \Rightarrow a + (m + 1 - 1)d = 2[a + (n + 1 - 1)d] \\
   \Rightarrow a + (m)d = 2[a + (n)d] \\
   \Rightarrow a + md = 2a + 2nd \\
   \Rightarrow md = a + 2nd \\
   \Rightarrow md - 2nd = a \\
   \Rightarrow d(m - 2n) = a \\
   \Rightarrow d = \dfrac{a}{{m - 2n}} \\
\]
Thus, the common difference \[d\] comes out to be \[\dfrac{a}{{m - 2n}}\].
Now, we first find the value of \[{(3m + 1)^{th}}\] term as
\[
   \Rightarrow t = a + (n - 1)d \\
   \Rightarrow {t_{3m + 1}} = a + (3m + 1 - 1)d \\
   \Rightarrow {t_{3m + 1}} = a + 3md \\
   \Rightarrow {t_{3m + 1}} = a + 3m\left( {\dfrac{a}{{m - 2n}}} \right) \\
   \Rightarrow {t_{3m + 1}} = a + \left( {\dfrac{{3ma}}{{m - 2n}}} \right) \\
   \Rightarrow {t_{3m + 1}} = \left( {\dfrac{{ma - 2na + 3ma}}{{m - 2n}}} \right) \\
   \Rightarrow {t_{3m + 1}} = \dfrac{{4ma - 2na}}{{m - 2n}} \\
\]
Thus, \[{t_{3m + 1}} = \dfrac{{4ma - 2na}}{{m - 2n}}\] which is the \[{(3m + 1)^{th}}\] term is equation (1).
Now, for \[{(m + n + 1)^{th}}\] term, its value can also be found out in a similar way.
\[
   \Rightarrow t = a + (n - 1)d \\
   \Rightarrow {t_{m + n + 1}} = a + (m + n + 1 - 1)d \\
   \Rightarrow {t_{m + n + 1}} = a + (m + n)d \\
   \Rightarrow {t_{m + n + 1}} = a + (m + n)\left( {\dfrac{a}{{m - 2n}}} \right) \\
   \Rightarrow {t_{m + n + 1}} = a + \left( {\dfrac{{ma + na}}{{m - 2n}}} \right) \\
   \Rightarrow {t_{m + n + 1}} = \left( {\dfrac{{ma - 2na + ma + na}}{{m - 2n}}} \right) \\
   \Rightarrow {t_{m + n + 1}} = \dfrac{{2ma - na}}{{m - 2n}} \\
\]
Thus, the value \[{(m + n + 1)^{th}}\] term is \[\dfrac{{2ma - na}}{{m - 2n}}\] , which will be denoted as equation (2).
Now if we divide equation (1) with equation (2), we get
 \[
     \Rightarrow \dfrac{{{t_{3m + 1}}}}{{{t_{m + n + 1}}}} = \dfrac{{\dfrac{{4ma - 2na}}{{m - 2n}}}}{{\dfrac{{2ma - na}}{{m - 2n}}}} \\
   \Rightarrow \dfrac{{{t_{3m + 1}}}}{{{t_{m + n + 1}}}} = \dfrac{{2(2ma - na)}}{{(2ma - na)}} \\
   \Rightarrow {t_{3m + 1}} = 2{t_{m + n + 1}} \\
\]
Thus, \[{(3m + 1)^{th}}\] term is twice that of \[{(m + n + 1)^{th}}\] term.

Note: In the given question, both the terms were of the same A.P. that’s why they had a common first term and common difference. It is important for a particular term in the subscript to avoid any confusion. Also it is necessary to be well versed with the common formulae of A.P.