. given that the nth term of the sequence is (3+n)/4, then the sum of the sequence to 105 terms will be
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we have nth term =(3+n)/4.
put n=1
first term=(3+1)/4=4/4=1
put n=2
second term =(3+2)/4=5/4
common difference = 5/4-1=1/4
sum of n terms =n/2[2a+(n-1)d]=105
n[2+(n-1)5/4]=210
n[2+5n/4-5/4]=210
n[3/4+5n/4]=210
3n/4+5n^2/4=210
5n^2 + 3n =210×4
5n^2 + 3n - 840 =0
now solve this quadratic equation and you will get the the correct value of n. hope this will help you. if you have any confusion comment.
put n=1
first term=(3+1)/4=4/4=1
put n=2
second term =(3+2)/4=5/4
common difference = 5/4-1=1/4
sum of n terms =n/2[2a+(n-1)d]=105
n[2+(n-1)5/4]=210
n[2+5n/4-5/4]=210
n[3/4+5n/4]=210
3n/4+5n^2/4=210
5n^2 + 3n =210×4
5n^2 + 3n - 840 =0
now solve this quadratic equation and you will get the the correct value of n. hope this will help you. if you have any confusion comment.
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