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Question
Q. First term is of arithmatic progression sequence is 73 and the difference between consecutive numbers is 9 , Find the sum numbers from 39 to 43 terms.
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Step : 1
Term at 39 position is 73 + (39 - 1)x 9
= [73 + (38)x 9 ]
= [73 + 342]
= [415]
ie. S39 = (415)
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Step : 2
Then,
Sum of 39 to 43 position is (5/2)[2*415 + (5 - 1)x 9 ]
= (2.5)[830 + (4)x 9]
= (2.5)[830 + 36]
= (2.5)[866]
ie. S39 to S43 = (2165)
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Step : 3
Second method:-
Term at 39 position is 73 + (39 - 1)x 9
= 415 (=t1)
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Step : 4
Term at 40 position is 73 + (40 - 1)x 9
= 424 (=t2)
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Step : 5
Term at 41 position is 73 + (41 - 1)x 9
= 433 (=t3)
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Step : 6
Term at 42 position is 73 + (42 - 1)x 9
= 442 (=t4)
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Step : 7
Term at 43 position is 73 + (43 - 1)x 9
= 451 (=t5)
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Step : 8
so the sequance of numbers from position 39 to 43 is 415, 424, 433, 442, 451
= 415 + 424 + 433 + 442 + 451
ie. S39 to S43 = 2165
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