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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).Find the ratio of their mth term.

Syeda

Answer.

Given ratio of sum of n terms of two APâ€™s = (7n+1):(4n+27)

Letâ€™s consider the ratio these two APâ€™s mth terms as am : aâ€™m â†’(2)

Recall the nth term of AP formula, an = a + (n â€“ 1)d

Hence equation (2) becomes,

am : aâ€™m = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

am : aâ€™m = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S2m â€“ 1 : Sâ€™2m â€“ 1

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Letâ€™s consider the ratio these two APâ€™s mth terms as am : aâ€™m â†’(2)

Recall the nth term of AP formula, an = a + (n â€“ 1)d

Hence equation (2) becomes,

am : aâ€™m = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

am : aâ€™m = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S2m â€“ 1 : Sâ€™2m â€“ 1

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Agam Gupta

Given ratio of sum of n terms of two APâ€™s = (7n+1):(4n+27)

Letâ€™s consider the ratio these two APâ€™s mth terms as am : aâ€™m â†’(2)

Recall the nth term of AP formula, an = a + (n â€“ 1)d

Hence equation (2) becomes,

am : aâ€™m = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

am : aâ€™m = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S2m â€“ 1 : Sâ€™2m â€“ 1

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Letâ€™s consider the ratio these two APâ€™s mth terms as am : aâ€™m â†’(2)

Recall the nth term of AP formula, an = a + (n â€“ 1)d

Hence equation (2) becomes,

am : aâ€™m = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

am : aâ€™m = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S2m â€“ 1 : Sâ€™2m â€“ 1

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Dinesh Singha

Given ratio of sum of n terms of two APâ€™s = (7n+1):(4n+27)

Letâ€™s consider the ratio these two APâ€™s mth terms as a_{m} : aâ€™_{m} â†’(2)

Recall the nth term of AP formula, a_{n} = a + (n â€“ 1)d

Hence equation (2) becomes,

a_{m} : aâ€™_{m} = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

a_{m} : aâ€™_{m} = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S_{2m â€“ 1} : Sâ€™_{2m â€“ 1 }

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Letâ€™s consider the ratio these two APâ€™s mth terms as a

Recall the nth term of AP formula, a

Hence equation (2) becomes,

a

On multiplying by 2, we get

a

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Priyanshu

Given ratio of sum of n terms of two APâ€™s = (7n+1):(4n+27)

Letâ€™s consider the ratio these two APâ€™s mth terms as am : aâ€™m â†’(2)

Recall the nth term of AP formula, an = a + (n â€“ 1)d

Hence equation (2) becomes,

am : aâ€™m = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

am : aâ€™m = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S2m â€“ 1 : Sâ€™2m â€“ 1

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Letâ€™s consider the ratio these two APâ€™s mth terms as am : aâ€™m â†’(2)

Recall the nth term of AP formula, an = a + (n â€“ 1)d

Hence equation (2) becomes,

am : aâ€™m = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

am : aâ€™m = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S2m â€“ 1 : Sâ€™2m â€“ 1

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Kishore Kumar

Given ratio of sum of n terms of two APâ€™s = (7n+1):(4n+27)

Letâ€™s consider the ratio these two APâ€™s mth terms as a_{m} : aâ€™_{m} â†’(2)

Recall the nth term of AP formula, a_{n} = a + (n â€“ 1)d

Hence equation (2) becomes,

a_{m} : aâ€™_{m} = a + (m â€“ 1)d : aâ€™ + (m â€“ 1)dâ€™

On multiplying by 2, we get

a_{m} : aâ€™_{m} = [2a + 2(m â€“ 1)d] : [2aâ€™ + 2(m â€“ 1)dâ€™]

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S_{2m â€“ 1} : Sâ€™_{2m â€“ 1 }

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

Letâ€™s consider the ratio these two APâ€™s mth terms as a

Recall the nth term of AP formula, a

Hence equation (2) becomes,

a

On multiplying by 2, we get

a

= [2a + {(2m â€“ 1) â€“ 1}d] : [2aâ€™ + {(2m â€“ 1) â€“ 1}dâ€™]

= S

= [7(2m â€“ 1) + 1] : [4(2m â€“ 1) +27] [from (1)]

= [14m â€“ 7 +1] : [8m â€“ 4 + 27]

= [14m â€“ 6] : [8m + 23]

Thus the ratio of mth terms of two APâ€™s is [14m â€“ 6] : [8m + 23].

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