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Find the sum of the first 25 terms of the A.P. whose 5th term is 34 and 9th term is 58. |
Answer» Given :-5th TERM is 34 9th term is 58 Solution :-a5 = a + ( n - 1)d = 34 a + ( 5 - 1 )d = 34 a + 4d = 34 .............eq( 1) a9 = a + ( 9 - 1 ) d = 58 a + 8d = 58 ..............eq( 2) Now , Subtract eq( 2) from eq( 1 )a + 8d - ( a + 4d ) = 58 - 34 a + 8d -a - 4d = 24 4d = 24 d = 24/4 d = 6 Hence , The common difference is 6 Now , put the value of d in eq( 1 )a + 4d = 34 a + 4 * 6 = 34 a + 24 = 34 a = 34 - 24 a = 10 The first term of an AP is 10 Now , The sum of first 25 TERMSSN = n/2 ( 2a + ( n - 1)d ) S25 = 25/2 ( 2 * 10 + ( 25 - 1) *6) S25 = 25/2 ( 20 + 144) S25 = 25/2 ( 164) S25 = 25 * 82 S25 = 2050 Hence , The sum of 25th terms of an AP is 2050 |
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