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

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.