Find the 4th term in AP whose sum is 20 and the sum of whose square is

1.	Find the 4th term in AP whose sum is 20 and the sum of whose square is 120.
Answer» Let the terms\xa0be\xa0a\xa0– 3d,\xa0a\xa0– d,\xa0a\xa0+ d, a+3dSum of the terms = (a\xa0– 3d) + (a\xa0– d) + (a\xa0+ d) + (a\xa0+ 3d) = 204a\xa0= 20a\xa0= 5Sum of the squares of the term = (a\xa0– 3d)2\xa0+ (a\xa0– d)2\xa0+ (a\xa0+ d)2\xa0+ (a\xa0+ 3d)2\xa0= 20a2\xa0– 6ad + 9d2\xa0+\xa0a2\xa0– 2ad + d2\xa0+\xa0a2\xa0+ 2ad + d2\xa0+\xa0a2\xa0+ 6ad + 9d2\xa0= 1204a2\xa0+ 20d2\xa0= 120 – – – – – – – – – – – – (a)Substituting\xa0a\xa0= 5 into (a)4(52) + 20d2\xa0= 120100 + 20d2\xa0= 120d =\xa0±\xa01d =\xa0± 1Thus, the four numbers are:Taking d = 1(a\xa0– 3d), (a\xa0– d), (a\xa0+ d), (a\xa0+ 3d) = (5 – 3), (5 – 1), (5 + 1), (5 + 3)\xa0= 2, 4, 6, 8Taking d = -1(a\xa0– 3d), (a\xa0– d), (a\xa0+ d), (a\xa0+ 3d)= (5 + 3), (5 + 1), (5 - 1), (5 - 3)= 8, 6, 4, 2

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