Question

19. The ratio of the sums of first m and first n terms of an arithmetic series is m n:
2 2
show that the ratio of the mth and nth terms is ^ ^h h

Answer 1

The ratio of the sum of m and n terms of an AP is m2 :n2 show that the ratio of the mth and nth term is ( 2m-1):(2n-1)

Solution :- Sm : Sn = m² : n²
Sm/Sn = m²/n²
We know, sum of r terms = Sr = r/2[2a + (r - 1) d]
Where, a is the first term and d is the common difference of an AP.

Now, Sm = m/2[2a + (m- 1)d]
Sn = n/2[2a + (n +1)d]
Now, Sm/Sn = m/2[2a + (m -1)d]/n/2[2a + (n -1)d]
m²/n² = m[2a + (m -1)d]/[2a + (n -1)d]
m/n = [2a + (m - 1)d ]/[2a + (n -1)d]
m[2a + (n - 1)d] = n[2a + (m - 1)d ]
2a(m - n) + d(mn - m - mn + n)= 0
2a(m - n) = d(m - n)
d = 2a ----(1)

Now, mth term = a + (m - 1)d
= a + (m - 1)2a = 2am - a = a(2m - 1) [ from equation (1)]
Similarly, nth term = a + (n -1)d
= a + (n -1)2a
= a + 2an - 2a = 2an - a = a(2n - 1)

Hence, mth term : nth term = (2m -1) : (2n -1)

Answer 2

Solution :

Let a be the first term and d

the common difference of an

A.P .

Then ,

The sums of m and n terms are

Sm = m/2[2a+(m-1)d] ----( 1 )

Sn = n/2[2a+(n-1)d] ------( 2 )

Sm/Sn = m²/n²

=> {m/2[2a+(m-1)d]}/{n/2[2a+(n-1)d]}

= m²/n²

=> [2a+(m-1)d]/[2a+(n-1)d] = m/n

=> [2a+(m-1)d]n = [2a+(n-1)d]m

=> 2a(n-m) = d[(n-1)m--(m-1)n]

=> 2a(n-m) = d(n-m)

=> d = 2a

Now ,

tm/tn = [a+(m-1)d]/[a+(n-1)d]

= [ a + ( m-1)2a ]/[a + ( n-1)2a ]

= ( 2m - 1 )/( 2n - 1 )

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