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.

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Vedantu Content Team · Accepted Answer

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.PComplete 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 asThe 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.

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.