Progressions - Arithmetic Progression
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Sequence P is an arithmetic progression with initial term 11 and common difference 3. Sequence Q is an arithmetic progression with third term 103 and common difference -4. To find the smallest value of k where the term pk is greater than qk+2, the expressions for the kth terms of P and Q are set up and solved for k. The smallest integer value of k that satisfies this is 15.
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Progressions - Arithmetic Progression
- 2. Qn: Arithmetic Progression (a) 6 (b) 11 (c) 14 (d) 15 Sequence P is defined by pn = pnâ1 + 3, p1 = 11, Sequence Q is defined as qn = qnâ1 â 4, q3 = 103. If pk > qk+2, what is the smallest value k can take?
- 3. Soln: Arithmetic Progression Sequence P is an A.P. with a = 11, and common difference 3. So, Pk = 11 + (k â 1)3. Sequence Q is an A.P. with third term 103 and common difference â 4. t3 = a + 2d 103 = a + 2 (â 4) or a = 111 qk + 2 = 111 + (k +1) (â 4) qk + 2 = 111 â 4k â 4 = 107 â 4k pk > qk+2 11 + (kâ1)3 > 107 â 4k Sequence P is defined by pn = pnâ1 + 3, p1 = 11, Sequence Q is defined as qn = qnâ1 â 4, q3 = 103. If pk > qk+2, what is the smallest value k can take?
- 4. Soln: Arithmetic Progression 8 + 3k > 107 â 4k 7k > 99 k > 99 7 k has to be an integer, so smallest value k can take is 15. Answer choice (d) Sequence P is defined by pn = pnâ1 + 3, p1 = 11, Sequence Q is defined as qn = qnâ1 â 4, q3 = 103. If pk > qk+2, what is the smallest value k can take?