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

Answer:

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)

Recall 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

Answer:

(14m - 6) : (8m + 23)

Step-by-step explanation:

First AP:

Let the first term be a and Common difference be d.

Sum of n terms = S(n) = (n/2)[2a + (n - 1) * d]

nth term = a(n) = a + (n - 1) * d

Second AP:

Let the first term be A and the common difference be D.

Sum of n terms = s(n) = n/2[2A + (n - 1) * D]

nth term = A(n) = A + (n - 1) * D

Ratio of mth term:

= [a + (m - 1) * d]/[A + (m - 1) * D]   ------ (i)

Given that sum of n terms of 1st AP/Sum of n terms of 2nd AP

$=>\frac{\frac{n}{2}[2a+(n-1)*d]}{\frac{n}{2}[2A+(n-1)*D]} = \frac{7n+1}{4n+27}$

$=>\frac{2a+(n-1)d}{2A+(n-1)D} =\frac{7n+1}{4n+27}$

$=>\frac{a+(\frac{n-1}{2})d}{A+(\frac{n-1}{2})D}=\frac{7n+1}{4n+27}$  

We need to find [a + (m - 1) * d]/[A + (m - 1) * D]

Hence, (n - 1)/2 = m - 1.

n = 2m - 1.

Putting n = 2m - 1 in above equation, we get

$=>\frac{a+(m-1)d}{A+(m-1)D}=\frac{7n+1}{4n+27}$

$=>\frac{a+(m-1)d}{A+(m-1)D}=\frac{7(2m-1)+1}{4(2m-1)+27}$

$=>\frac{a+(m-1)d}{A+(m-1)D}=\frac{14m-7+1}{8m-4+27}$

$=>\frac{a+(m-1)d}{A+(m-1)D}=\frac{14m-6}{8m+23}$

Therefore, Ratio of mth terms = (14m - 6) : 8m + 23.

Hope it helps!