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Question
Q. First term is of arithmatic progression sequence is 79 and the difference between consecutive numbers is 13 , Find the sum numbers from 44 to 48 terms.
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Step : 1
Term at 44 position is 79 + (44 - 1)x 13
= [79 + (43)x 13 ]
= [79 + 559]
= [638]
ie. S44 = (638)
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Step : 2
Then,
Sum of 44 to 48 position is (5/2)[2*638 + (5 - 1)x 13 ]
= (2.5)[1276 + (4)x 13]
= (2.5)[1276 + 52]
= (2.5)[1328]
ie. S44 to S48 = (3320)
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Step : 3
Second method:-
Term at 44 position is 79 + (44 - 1)x 13
= 638 (=t1)
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Step : 4
Term at 45 position is 79 + (45 - 1)x 13
= 651 (=t2)
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Step : 5
Term at 46 position is 79 + (46 - 1)x 13
= 664 (=t3)
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Step : 6
Term at 47 position is 79 + (47 - 1)x 13
= 677 (=t4)
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Step : 7
Term at 48 position is 79 + (48 - 1)x 13
= 690 (=t5)
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Step : 8
so the sequance of numbers from position 44 to 48 is 638, 651, 664, 677, 690
= 638 + 651 + 664 + 677 + 690
ie. S44 to S48 = 3320
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