Find 17th term of an AP whose 4th term is 13 and the 10th term is 25 .
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Simplification of the above question cane written as
Simplification of the above question cane written asa+ 3d = 13
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting(1) in (2)
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting(1) in (2)3(13 - a) = 25 - a
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting(1) in (2)3(13 - a) = 25 - a39 - 3a = 25 - a
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting(1) in (2)3(13 - a) = 25 - a39 - 3a = 25 - a2a. = 14
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting(1) in (2)3(13 - a) = 25 - a39 - 3a = 25 - a2a. = 14a = 7
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting(1) in (2)3(13 - a) = 25 - a39 - 3a = 25 - a2a. = 14a = 7From (1) d= 2
Simplification of the above question cane written asa+ 3d = 13=>3d = 13 - a. (1)a+ 9d = 25=> 9d = 25- a3(3d) = 25 - a. (2)Substituting(1) in (2)3(13 - a) = 25 - a39 - 3a = 25 - a2a. = 14a = 7From (1) d= 217th term = 7 + 15(2) =37 the answer
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