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if the sum of n terms of an A.P. is 3n2\xa0+ 5n and its mth\xa0 term is 164, find the value of m |
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Answer» It is given that sum of n terms of an A.P. is 3n2\xa0+ 5n=> S1\xa0= 3(1)2\xa0+ 5(1) = 3 + 5 = 8 which will also be 1st term as "a1".Further, S2\xa0\xa0= 3 (2)2\xa0+ 5 (2) = 12 + 10 = 22And, we know, a2\xa0=\xa0S2\xa0-\xa0S1\xa0= 22 - 8 = 14....\xa0So, d =\xa0a2\xa0- a1\xa0= 14 - 8 = 6Now, it\'s given that am\xa0= 164 =>\xa0a1\xa0+ (m - 1)d = 1648 +\xa0(m - 1)×6 = 164 =>\xa0(m - 1)×6 = 156\xa0=>\xa0(m - 1) = 26\xa0=> m = 27 (Answer)\xa0 According to question,S1 = a = 3(1)2 + 5 x 1 = 8S2 =\xa03(2)2 + 5 x\xa02 = 22 Then, a2 = S2 - S1 = 22 - 8 = 14S3 =\xa03(3)2 + 5 x\xa03 =\xa042 Then, a3 = S3 - S2 =\xa042 -\xa022 = 22Therefore, d = 14 - 8 = 6Now, am = a + (m - 1)d = 164=> 164 = 8 + (m - 1)6=> 156 = (m - 1)6=> 26 = m - 1=> m = 27 |
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