If the ratio of the sums of $m$ and $n$ terms of an A.P. is $m^2 : n^2$, show that the ratio of its $m$th and $n$th terms is $(2m-1) : (2n-1)$.

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> [!TLDR] Starting from the given ratio of sums $S_m:S_n=m^2:n^2$, algebraic simplification forces the relation $d=2a$ between the common difference and first term; substituting this back into the $m$th and $n$th term formulas proves the required ratio.

> [!FORMULA] For an A.P. with first term $a$ and common difference $d$:
$$S_n = \frac{n}{2}\big[2a+(n-1)d\big], \qquad a_n = a+(n-1)d$$

1. **Write the given ratio** using the sum formula for $S_m$ and $S_n$:
   $$\frac{S_m}{S_n} = \frac{\frac{m}{2}\big[2a+(m-1)d\big]}{\frac{n}{2}\big[2a+(n-1)d\big]} = \frac{m^2}{n^2}$$
2. **Cancel the common factor $	frac12$** top and bottom:
   $$\frac{m\big[2a+(m-1)d\big]}{n\big[2a+(n-1)d\big]} = \frac{m^2}{n^2}$$
3. **Cross-multiply:**
   $$n^2\cdot m\big[2a+(m-1)d\big] = m^2\cdot n\big[2a+(n-1)d\big]$$
4. **Divide both sides by $mn$** (both positive, since $m,n$ are term-counts):
   $$n\big[2a+(m-1)d\big] = m\big[2a+(n-1)d\big]$$
5. **Expand both sides:**
   $$2an+n(m-1)d = 2am+m(n-1)d$$
6. **Bring like terms together:**
   $$2an-2am = m(n-1)d - n(m-1)d$$
   $$2a(n-m) = d\big[mn-m-mn+n\big] = d(n-m)$$
7. **Since $m
e n$, divide both sides by $(n-m)$:**
   $$2a = d \quad	ext{i.e.}\quad d = 2a$$
8. **Now compute the ratio of the $m$th and $n$th terms**, substituting $d=2a$:
   $$a_m = a+(m-1)d = a+(m-1)(2a) = a\big[1+2(m-1)\big] = a(2m-1)$$ …

Q.If the ratio of the sums of $m$ and $n$ terms of an A.P. is $m^2 : n^2$ , show that the ratio of its $m$ th and $n$ th terms is $(2m-1) : (2n-1)$ .

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Why This Matters for Exams

Q.If the ratio of the sums of mmm and nnn terms of an A.P. is m2:n2m^2 : n^2m2:n2, show that the ratio of its mmmth and nnnth terms is (2m−1):(2n−1)(2m-1) : (2n-1)(2m−1):(2n−1).

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Q.If the ratio of the sums of $m$ and $n$ terms of an A.P. is $m^2 : n^2$ , show that the ratio of its $m$ th and $n$ th terms is $(2m-1) : (2n-1)$ .

Unlock everything free for 14 days