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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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Answered by Shadow1357

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Answered by naitik1990

Let S

Let S m

Let S m and S

Let S m and S n

Let S m and S n 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

Let S m and S n 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 differenceS

Let S m and S n 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 differenceS n

Let S m and S n 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 differenceS n 2

Let S m and S n 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 differenceS n 2 m

Let S m and S n 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 differenceS n 2 m 2

2n [2a+(n−1)d]

2n [2a+(n−1)d]2

2n [2a+(n−1)d]2m

2n [2a+(n−1)d]2m [2a+(m−1)d]

2n [2a+(n−1)d]2m [2a+(m−1)d] n

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d n

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d nm

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d nm n[2a+(m−1)d]=m[2a+(n−1)d]

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d nm n[2a+(m−1)d]=m[2a+(n−1)d]2an+mnd−nd+2am+mnd−nd

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d nm n[2a+(m−1)d]=m[2a+(n−1)d]2an+mnd−nd+2am+mnd−ndmd−nd=2am−2an

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d nm n[2a+(m−1)d]=m[2a+(n−1)d]2an+mnd−nd+2am+mnd−ndmd−nd=2am−2an(m−n)d=2a(m−n)

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d nm n[2a+(m−1)d]=m[2a+(n−1)d]2an+mnd−nd+2am+mnd−ndmd−nd=2am−2an(m−n)d=2a(m−n)d=2a

2n [2a+(n−1)d]2m [2a+(m−1)d] n 2 m 2 2a+(n−1)d2a+(m−1)d nm n[2a+(m−1)d]=m[2a+(n−1)d]2an+mnd−nd+2am+mnd−ndmd−nd=2am−2an(m−n)d=2a(m−n)d=2aNow, the ratio of mth and nth terms is

a+(n−1)d

a+(n−1)da+(m−1)d

a+(n−1)da+(m−1)d a+(n−1)2a

a+(n−1)da+(m−1)d a+(n−1)2aa+(m−1)2a

a+(n−1)da+(m−1)d a+(n−1)2aa+(m−1)2a a(1+2n−2)

a+(n−1)da+(m−1)d a+(n−1)2aa+(m−1)2a a(1+2n−2)a(1+2m−2)

a+(n−1)da+(m−1)d a+(n−1)2aa+(m−1)2a a(1+2n−2)a(1+2m−2) 2n−1

2m-1

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

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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).✔Need Proper explained answer.✖ Don't spam.​

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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).

✔Need Proper explained answer.

✖ Don't spam.