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If the sum of first 7 terms aof an A.P. is 10 and sum of next 7 terms is 17 THEN find the A.P. |
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Answer» Let first term of AP = aCommon difference of AP = dSum of first 7 terms = 10So,\xa0{tex}10={7\\over 2}[2a+6d]\\\\=> 20 = 14a+42d \\ .......(1){/tex}Sum of next 7 terms = sum of 14 terms - sum of first 7 terms{tex}17 = {14\\over 2}[2a+13d]-10\\\\=> 27 = 14a+91d\\ \\ .....(2){/tex}Subtract (1) from (2), we get=> 7 = 49d=> d =\xa0{tex}1\\over 7{/tex}Put value of d in (1), we get=>\xa0{tex}20 = 14a + 6\\\\=> 14a = 14\\\\=> a = 1{/tex}AP :\xa0{tex}1, {8\\over 7},{15\\over 7},{22\\over 7},.....{/tex}\xa0\xa0 sum of first 7 terms = 10sum of next 7 terms = 17sum of first 14 terms = (10 + 17) = 27let first term = a and common difference = dthen S7\xa0= n/2(2a + (n = 1)d 10 = 7/2(2a + (7 - 1)d ............. 10 = 7a + 21d ........ ialso S14\xa0= 14/2 (2a + (14 - 1)d ...........27 = 14a + 91d ...... ii from i and ii we get 7 = 49d ,........ d = 1/710\xa0= 7a + 3 ................................... a = 1Thus the AP is defined by a , a+d , a+2d, ................. 1 , 8/7 , 9/7, ...............\xa0 |
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