Arithmetic Series ::
An Arithmetic Series is a series of numbers in which each term increases by a constant amount.
How to find the sum of the Arithmetic Sequence or Series for the given Series ::
When the series contains a large amount of numbers, its impractical to add manually. You can quickly find the sum of any arithmetic sequence by multiplying the average of the first and last term by the number of terms in the sequence.
That is given by Where n = number of terms, a1 = first term, an = last term
Here Last term is given by where d = common difference
Now given Arithmetic Series is
2, 5, 8, 11,...
Here a1 = 2, d = 3, n = 36
Now,
Now, Sum to 36 terms is given by
Therefore, Sum to 36 terms of the series 2, 5, 8, 11,... is 1962.
0.0169 / 0.0130 = 169 / 130
= 13 / 10
Given Exp. = 4 / 7 + {(2q - p) / (2q + p)}
Dividing numerator as well as denominator by q,
Exp = 4/7 + {2-p/q) / (2 + p/q)}
= 4/7 + {(2 - 4/5) / (2 + 4/5)}
= 4/7 + 6/14
= 4/7 + 3/7
=7/7
=1.
25% of 25% = (25/100) x (25/100) = 625/10000 = 0.625
Let y% of 20 = .05
Then, (y x 20)/100 = .05
? y = .25
2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
81/3% = (25/3 x 1/100) = 1/12
.025 = (25/1000) x 100% = 2.5%
Let N% of 2/7 is 1/35
? (2/7 x N) / 100 = 1/35
? N = 1/35 x 7/2 x 100 = 10%
Let the money interest at 8% interest be ? P .
Then, the money interest at 10% interest = ?(4000 - P)
According to the question,
(P x 8 x 1)/100 + [(4000 - P) x 10 x 1]/100 = 352
? 8P + 40000 - 10P = 35200
? 40000 - 35200 = 2P
? P = 4800/2 = ? 2400
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