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If nnn-th term of an AP series is 5n−25n - 25n−2, then sum of nnn terms is:"
Possible options: A) 510(5n2−n)\frac{5}{10}(5n^2 - n)105(5n2−n) B) 510(5n2+n)\frac{5}{10}(5n^2 + n)105(5n2+n) C) 510(5n2−2n)\frac{5}{10}(5n^2 - 2n)105(5n2−2n) D) 510(5n2+2n)\frac{5}{10}(5n^2 + 2n)105(5n2+2n)?
Possible options: A) 510(5n2−n)\frac{5}{10}(5n^2 - n)105(5n2−n) B) 510(5n2+n)\frac{5}{10}(5n^2 + n)105(5n2+n) C) 510(5n2−2n)\frac{5}{10}(5n^2 - 2n)105(5n2−2n) D) 510(5n2+2n)\frac{5}{10}(5n^2 + 2n)105(5n2+2n)?
Most Upvoted Answer
If nnn-th term of an AP series is 5n−25n - 25n−2, then sum of nnn term...
Given:
The n-th term of an AP series is 5n−2.
Formula for n-th term of an AP series:
The n-th term of an AP series is given by the formula:
an = a1 + (n-1)d
where:
an = n-th term of the AP series
a1 = first term of the AP series
d = common difference between consecutive terms
n = number of terms
Given n-th term:
In this case, the n-th term of the AP series is given as 5n−2.
So, we can write:
5n−2 = a1 + (n-1)d
=> 5n−2 = a1 + dn−d
=> 5n−2 = a1 + dn−d
Finding the sum of n terms:
To find the sum of n terms of an AP series, we use the formula:
Sn = n/2[2a1 + (n-1)d]
Comparing with given n-th term:
By comparing the given n-th term (5n−2) with the formula for the n-th term of an AP series, we can find the values of a1 and d.
a1 = -2
d = 5
Substituting in the sum formula:
Now, we substitute the values of a1 and d in the sum formula to find the sum of n terms.
Sn = n/2[2(-2) + (n-1)(5)]
=> Sn = n/2[-4 + 5n - 5]
=> Sn = n/2[5n - 9]
Answer:
Therefore, the sum of n terms of the given AP series is 5n^2 - 9n.
So, the correct option is:
A) 5/10(5n^2 - n)
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If nnn-th term of an AP series is 5n−25n - 25n−2, then sum of nnn terms is:"Possible options: A) 510(5n2−n)\frac{5}{10}(5n^2 - n)105(5n2−n) B) 510(5n2+n)\frac{5}{10}(5n^2 + n)105(5n2+n) C) 510(5n2−2n)\frac{5}{10}(5n^2 - 2n)105(5n2−2n) D) 510(5n2+2n)\frac{5}{10}(5n^2 + 2n)105(5n2+2n)?
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