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Contents
Arithmetic Progressions 1 2
| Arithmetic Progressions   Thesum of the first 10 terms in an arithmetic progression is 50 and the sum of thenext 10 terms is 250. Find the thirteenth term. Solution: Sum of first 10terms = 10/2 [2a + (10 - 1) d] = 50 5 (2a + 9d) = 50 2a + 9d = 10 .....(1) Sum of first 20terms = 20/2 [2a + (20 - 1) d] = 250 + 50 10 (2a + 19d) = 300 2a + 19d = 30 .....(2) Solving (1) and(2) simultaneously, d = 2, a = -4 13th term = a +(13 - 1) d = -4 + 12(2) = 20 Thesecond term of an arithmetic progression is nine times the fifth term and thesum of the first eight terms is 56. Find (i) the first term and common difference. (ii) the least number of terms of the A. P. which must be taken for the sum to be negative. Solution: Second term = 9 (Fifth term) a + d = 9 (a + 4d) 8a + 35d =0 .....(1) Sum to first eight terms: 8/2 [2a + (8 - 1) d] = 56 8a + 28d = 56 .....(2) Solving (1) and (2) simultaneously, d = -8, a = 35 Let the leastnumber of terms be n. n/2 [2 (35) + (-8) (n - 1)] < 0 39n - 4n2 < 0 n (39 - 4n) < 0 n > 9 3/4 sincen > 0 The least number ofterms is 10. Theseries lg x + lg 2 + lgx2 + lg 4 + lgx3 + lg 8 + ... is an arithmetic progression.Show that the sum of first ten terms is 55 lg 2x. Solution: lg x + lg 2 + lgx2 + lg 4 + lgx3 + lg 8 + ... = (lg x + lg 2) + 2 (lg x + lg 2) + 3 (lg x + lg 2) + ... powerlaw of logarithms This is an A. P.with first term = lg x + lg 2 = lg 2x common difference = lg x + lg 2 = lg 2x Sum to first 10terms = 10/2 [ 2 lg 2x + (10 - 1) lg 2x] = 5 (11 lg 2x) = 55 lg 2x   |
