a man saves rupees 32 during 1st year, rupees 36 in next year and rupees 40 in 3rd year if he continues his savings in a sequence in how many years will he save rupees 2000 ?
Answers
Answered by
2
a=32,d=4
2000=n/2(2×32+(n-1)×4)
4000=n(64+4n-4)
4000=60n+4n^2
n^2+15n-1000=0
solve it
2000=n/2(2×32+(n-1)×4)
4000=n(64+4n-4)
4000=60n+4n^2
n^2+15n-1000=0
solve it
Answered by
2
Since it is in ap
a=32
d=4
Sn=2000
Sn=n/2[2a-(n-1)d]
Substituting the value we get the value of n
Hope it helps
a=32
d=4
Sn=2000
Sn=n/2[2a-(n-1)d]
Substituting the value we get the value of n
Hope it helps
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