Question

Answer the following:
4x4 = 16
1. The ratio of the sums of m and n terms of an A.P.is m:n. Show that the ratio
of the mand nth terms is (2m-1): (2n-1).​

Answer 1

Correct Question -

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

Show that the ratio of the m and n th terms is (2m-1) : (2n-1).

Solution -

In the above Question , the following information is given -

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

To Prove -

Show that the ratio of the m and n th terms is (2m-1) : (2n-1).

Solution -

We know that -

Sum of n terms in an AP

=> $\sf{ \dfrac{ n }{ 2 } ( 2a + ( n - 1 ) d ) }$

Where,

n refers to the number of terms .

a is the innitial term .

d is the Common difference .

So ,

Sum of m terms of an ap -

=> $\sf{ \dfrac{ m }{ 2 } ( 2a + ( m - 1 ) d ) }$ ........ { 1 }

Sum of n terms of an ap -

=> $\sf{ \dfrac{ n }{ 2 } ( 2a + ( n - 1 ) d ) }$ ......... { 2 }

The ratio of { 1 } and { 2 } is m : n

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

$\sf{ \implies { \dfrac{ ( 2a + ( m - 1 ) d ) }{ ( 2a + ( n - 1 ) d ) } = \dfrac{m}{n} }}$

Now , we also know that -

n th term of an ap -

$\sf{ a_n = a + ( n - 1 ) d }$

Where -

a_n is the nth term.

a Is the innitial term .

d is th common difference .

n is the number of terms.

So ,

Nth term -

$\sf{ a + ( n - 1 ) d }$

Nth term -

$\sf{ a + ( m - 1 ) d }$

Ratio of mth term / n the term -

$\sf { \dfrac{ ( a + ( m - 1) d ) }{ ( a + ( n - 1 ) d ) } }$

Now , we will use this result .

$\sf{ \implies { \dfrac{ ( 2a + ( m - 1 ) d ) }{ ( 2a + ( n - 1 ) d ) } = \dfrac{m}{n} }}$

Now ,

Substitute m as 2m - 1 and n as 2n - 1

$\sf{ \implies { \dfrac{ ( 2a + ( 2m - 2 ) d ) }{ ( 2a + ( 2n - 2 ) d ) } = \dfrac{2m - 1}{2n - 1} }}$

Now , cancel 2

$\sf{ \implies { \dfrac{ ( a + ( m - 1) d ) }{ ( a + ( n - 1 ) d ) } = \dfrac{2m - 1}{2n - 1} }}$

Hence Proved .

Note that the given Values are wrong .

If the ratio of the sums of m and n terms of an A.P.is m : n , it is not possible for the ratio of the mand nth terms to be 2m - 1 : 2 n - 1

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