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Venkatesh

Nov 13, 2013

The ratio of the sum of n terms of two A.P.s is (7n+1):(4n+27).

The ratio of the sum of n terms of two A.P.s is (7n+1):(4n+27).Find the ratio of their mth term.

Answers(6)

Answer

Syeda

Answer.
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].

Agam Gupta

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].

Dinesh Singha

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].

Priyanshu

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].

Kishore Kumar

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].

STAFF-95393

\\Solution\\ Sum\ of\ n\ the\ terms\ in\ 1^{st}\ AP=\frac{n}{2}(2a_1+(n-1)d_1)\\\\ Sum\ of\ n\ the\ terms\ in\ 2^{nd}\ AP=\frac{n}{2}(2a_2+(n-1)d_2)\\ Given:\\ Ratio\ of\ n\ terms\ of\ two\ AP's=\frac{7n+1}{4n+27}\\\\ \Rightarrow \frac{\frac{n}{2}(2a_1+(n-1)d_1)}{\frac{n}{2}(2a_2+(n-1)d_2)}=\frac{7n+1}{4n+27}\\\\\\ \Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}\\\\\\ \Rightarrow \frac{a_1+(\frac{n-1}{2})d_1}{a_2+(\frac{n-1}{2})d_2}=\frac{7n+1}{4n+27}\ \ \ \ \rightarrow (1)\\\\\\ Let,\ \frac{n-1}{2}=m-1\\\\ \Rightarrow n-1=2(m-1)\\\\ \Rightarrow n=2m-2+1\\\\ \Rightarrow n=2m-1\\\\ Substitute\ above\ values\ in\ equation\ (1)\\\\ \Rightarrow \frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{7(2m-1)+1}{4(2m-1)+27}\\\\\\ \Rightarrow \frac{T_{m1}}{T_{m2}}=\frac{14m-7+1}{8m-4+27}=\frac{14m-6}{8m+23}\\\\\\ \therefore Required\ m^{th}\ terms\ ratio\ of\ two\ AP's=\frac{14m-6}{8m+23}
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