How many terms of AP 5,9,13,....will give a sum of 275?
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Sum of n terms = 275 | common difference = 4
n/2(10+(n-1)4) = 275 |
2n^2 + 3n = 275
n= 11 it is +ve is accepted term
n/2(10+(n-1)4) = 275 |
2n^2 + 3n = 275
n= 11 it is +ve is accepted term
shivmking74:
wrong answer dude
it's right
Answered by
2
Sn = n/2 {2a + (n-1)d}
275 = n/2 {2*5+(n-1)4}
550=n(10+4n-4)
550 =n(4n-6)
550=4n^2+6n
4n^2+6n =550
D=b^2-4ac
D= 6^2-4*4*-550
D =8836
n= -b+-√D/2a
n= -6+-√8836/2*4
n=11
negative term Will be neglect because term never be negative
Hope it's help you
275 = n/2 {2*5+(n-1)4}
550=n(10+4n-4)
550 =n(4n-6)
550=4n^2+6n
4n^2+6n =550
D=b^2-4ac
D= 6^2-4*4*-550
D =8836
n= -b+-√D/2a
n= -6+-√8836/2*4
n=11
negative term Will be neglect because term never be negative
Hope it's help you
correct answer
how it can be correct it's is a quadratic equation it should have two roots not one
yes this question has 2 roots but negative answer neglected
yeah ✌
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