Question

If the ratio of sum of first $n$ terms of two A.P's is $(7n+1)\:\colon(4n+27)$, find the ratio of their $m^{th}$ terms.

Answer 1

Solution:

Step-by-step explanation:

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)

Let’s consider the ratio these two AP’s mth terms as am : a’m →(2)

The nth term of AP formula, an = a + (n – 1)d

Hence equation (2) becomes,

am : a’m = a + (m – 1)d : a’ + (m – 1)d’

On multiplying by 2, we get

am : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]

= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]

= S2m – 1 : S’2m – 1

= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]

= [14m – 7 +1] : [8m – 4 + 27]

= [14m – 6] : [8m + 23]

Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].

Answer 2

Sn1/Sn2 = n/2 ( 2a + ( n-1)d)/ ( n/2) ( 2 a1 + ( n-1) d1)

= 2a + (n-1)d )/ 2a1 + ( n-1)d1 = 7n +1)/ 4n +27

2a + nd -d)/ 2a1 + nd1 -d1 = 7n+1)/ 4n+27

2a -d + nd)/2a1-d1 +nd1 = 7n +1)/ 4n +27

Compare

2a - d = 1

d = 7

2a = 1 +d = 8

a= 4

Also 2a1 -d1 = 27

d1 = 4

2a1 = 31

a1 = 31/2



am/am1 = a + ( m-1)d/ a1 + ( m-1) d1

= 4 + ( m-1)7/ 31/2 + ( m-1)4

= 4 + 7m -7 )/ 31 + 8m -8)/2

= -3 + 7m)2)/ 8m +23

= -6 + 14m)/ 8m +23