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From Karnataka Open Educational Resources
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, 06:57, 8 January 2013
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| 1. When we combine Algebra and Geometry - this feature is not as good in CaRMetal. | | 1. When we combine Algebra and Geometry - this feature is not as good in CaRMetal. |
| 2. When we combine statistics (spreadsheets) , with algebra, charts - Geometry - this feature is not there in CaRMetal. | | 2. When we combine statistics (spreadsheets) , with algebra, charts - Geometry - this feature is not there in CaRMetal. |
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| + | =P th term of an AP is Q and Q th term of an AP is P . Then find PQ th term ?= |
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| + | '''Mallikarjun Sudi, Ghs Yelhari''' All maths teachers pls solve this problem |
| + | P th term of an AP is Q and Q th term of an AP is P . Then find PQ th term ? |
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| + | ''' Sneha Titus, University Resource Centre, Azim Premji University''' |
| + | This is a very nice problem. Here is the solution that I worked out. |
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| + | Let the nth term be Tn = a + (n‒1)d, where a is the first term and d is the common difference. |
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| + | Given: Tp = q = a + (p‒1)d |
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| + | And Tq = p = a + (q‒1)d |
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| + | So Tp ‒Tq = q ‒ p = a + (p‒1)d ‒[a + (q‒1)d] |
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| + | q ‒ p = pd ‒ d ‒ qd + d |
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| + | q ‒ p = (p ‒ q) d |
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| + | ∴ d = = ‒ 1 |
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| + | And since Tq = p = a + (q‒1)d, |
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| + | p = a + (q ‒1)(‒1) = a ‒q + 1 |
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| + | so that a = p + q ‒1 |
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| + | Now, Tpq = a + (pq‒1)d = a + (pq – 1)(‒1) = a ‒pq + 1 = p + q ‒1 ‒pq + 1 = p + q ‒pq |
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| + | Example T2 = 4 and T4 = 2, what is T8 |
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| + | T2 = 4 = a + (2‒1)d = a + d |
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| + | T4 = 2 = a + (4‒1)d = a + 3d |
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| + | T2 ‒T4 = 4 ‒ 2 = a + d ‒[a + 3d] = – 2d |
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| + | 2 = ‒2d |
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| + | ∴ d = ‒1 |
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| + | And T2 = 4 = a + (2‒1)(‒1) = a ‒1 |
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| + | ∴ a = 4 + 1 = 5 |
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| + | And T8 = a + (8‒1)d] = a +7 d = a ‒ 7 = 5 ‒ 7 = ‒2 |
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| + | The A.P with a = 5, d = ‒1 is |
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| + | 5, 4, 3, 2, 1, 0, ‒1, ‒2,………. |
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| + | Notice that T2 = 4 and T4 = 2 and T8 = ‒2 |
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| = Which day is Pi Day ? = | | = Which day is Pi Day ? = |