Question

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The ratio of the sums of m and n terms of an A.P. is m²: n². Show that the ratio of mth and nth term is (2m – 1): (2n – 1).

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Answer 1

Answer:

Let Sm and Sn be the sum of the first m and first n terms of the AP respectively. Let, a be the first term and d be a common difference

SnSm=n2m2

⇒2n[2a+(n−1)d]2m[2a+(m−1)d]=n2m2

⇒2a+(n−1)d2a+(m−1)d=nm

⇒n[2a+(m−1)d]=m[2a+(n−1)d]

⇒2an+mnd−nd+2am+mnd−nd

⇒md−nd=2am−2an

⇒(m−n)d=2a(m−n)

⇒d=2a

Now, the ratio of mth and nth terms is 

an/am

=a+(n−1)da+(m−1)d=a+(n−1)2aa+(m−1)2a

                                       =a(1+2n−2)a(1+2m−2)

                                       =2n−12m−1

Thus, ratio of its mth and nth terms is 2m−1:2n−1

Step-by-step explanation:

Hope it helps you

Answer 2

Question :

The ratio of the sums of m and n terms of an A.P. is m²: n². Show that the ratio of mth and nth term is (2m – 1): (2n – 1).

Solution :

Given, the ratio of the sums of m and n terms of an A.P. is m²: n².

We have to show ; Show that the ratio of mth and nth term is (2m – 1): (2n – 1).

Now, we know,

Sum of nth terms in an A.P. = n/2 [2a + (n - 1)d]

∴ Sum of mth terms in an A.P. = m/2 [2a + (m - 1)d]

Now atq,

$\longmapsto\sf \dfrac{m}{2} [2a + (m - 1)d] : \dfrac{n}{2} [2a + (n - 1)d] = m^2 : n^2$

$\longmapsto\sf \dfrac{ \dfrac{m}{2} + [2a + (m - 1)d]}{ \dfrac{n}{2} + [2a + (n - 1)d]} = \dfrac{m^2}{n^2} \\\\\ \\ \longmapsto\sf \dfrac{m + [2a + (m - 1)d]}{n + [2a + (n - 1)d]}= \dfrac{m^2}{n^2} \\\\ \\ \longmapsto\sf \dfrac{n}{m} \times \dfrac{m + [2a + (m - 1)d]}{n + [2a + (n - 1)d]} = \dfrac{m^2}{n^2} \times \dfrac{n}{m} \\\\ \\ \longmapsto\sf \dfrac{2a + (m - 1)d}{2a + (n - 1)d} = \dfrac{m}{n} \\\\ \\ \longmapsto\sf n \left\{2a + (m -1)d\right\} = m \left\{2a + (n - 1)d \right\}$

$\longmapsto\sf n\left\{2a + md - d\right\} = m \left\{2a + nd - d\right\} \\\\ \\ \longmapsto\sf 2an + mnd - nd = 2am + mnd - md \\\\ \\ \longmapsto\sf 2an - 2am + mnd - mnd - nd + md = 0 \\\\ \\ \longmapsto\sf 2a(n - m) -d(n - m) = 0 \\\\ \\ \longmapsto\sf 2a(n - m) = d(n - m) \\\\ \\ \longmapsto\sf 2a = d \qquad\qquad[Dividing\ both \ sides\ by\ (n - m)] \\\\ \\ \longmapsto\underline{\boxed{\bold{d = 2a}}}$

Now we know that,

nth term of an A.P. = a + (n - 1)d

∴ mth term of an A.P. = a + (m - 1)d

Now ratio of mth term and nth term :

$\longmapsto\sf \left\{a + (m - 1)d\right\} : \left\{a + (n - 1)d\right\} \\\\ \\ \longmapsto\sf \dfrac{\left\{a + (m - 1)d\right\}}{\left\{a + (n - 1)d\right\}} \\\\ \\ \longmapsto\sf \dfrac{a + (m - 1)2a}{a + (n - 1)2a} \\\\ \\ \longmapsto\sf \dfrac{a + 2am - 2a}{a + 2an - 2a} \\\\ \\ \longmapsto\sf \dfrac{a(1 + 2m - 2)}{a(1 + 2n - 2)} \\\\ \\ \longmapsto\sf \dfrac{2m - 1}{2n - 1} \\\\ \\ \longmapsto\underline{\boxed{\bold{(2m - 1) : (2n - 1)}}}$

Therefore,

Ratio of mth and nth term = (2m - 1) : (2n - 1)  [Showed]