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Question
Q. First term is of arithmatic progression sequence is 73 and the difference between consecutive numbers is 13 , Find the sum numbers from 34 to 38 terms.
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Step : 1
Term at 34 position is 73 + (34 - 1)x 13
= [73 + (33)x 13 ]
= [73 + 429]
= [502]
ie. S34 = (502)
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Step : 2
Then,
Sum of 34 to 38 position is (5/2)[2*502 + (5 - 1)x 13 ]
= (2.5)[1004 + (4)x 13]
= (2.5)[1004 + 52]
= (2.5)[1056]
ie. S34 to S38 = (2640)
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Step : 3
Second method:-
Term at 34 position is 73 + (34 - 1)x 13
= 502 (=t1)
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Step : 4
Term at 35 position is 73 + (35 - 1)x 13
= 515 (=t2)
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Step : 5
Term at 36 position is 73 + (36 - 1)x 13
= 528 (=t3)
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Step : 6
Term at 37 position is 73 + (37 - 1)x 13
= 541 (=t4)
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Step : 7
Term at 38 position is 73 + (38 - 1)x 13
= 554 (=t5)
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Step : 8
so the sequance of numbers from position 34 to 38 is 502, 515, 528, 541, 554
= 502 + 515 + 528 + 541 + 554
ie. S34 to S38 = 2640
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