In an AP, the sum of first n terms is 3n^2+13n/2   .Find the 25th term.
qais:
actually it's (3n^2)/2 + 13n/2
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Answered by
4
(3*25^2+13*25/2)-(3*24^2+13*24/2)
=> (5625+212.5)-(5184+156)
=> 5837.5-5340
=497.5
=> (5625+212.5)-(5184+156)
=> 5837.5-5340
=497.5
Answered by
16
Actually it's 3n²/2 + 13n/2. considering this only.
n term sum = 3n²/2 + 13n/2
as we know that nth term = (Sum of nth term ) -( sum of (n-1)th term)
method 1
so 25th term = (sum of 25 terms) - ( sum of 24 terms)
=(3×(25)²/2 +(13×25)/2) - ( 3×(24)²/2 + (13×24)/2)
= 1100 - 1020
= 80
method 2
1st term = 8
sum of 2 terms = 19
2nd term = 19 - 8= 11
so. difference = 11- 8 = 3
25th term = (8 + (25 -1)×3) = 80
n term sum = 3n²/2 + 13n/2
as we know that nth term = (Sum of nth term ) -( sum of (n-1)th term)
method 1
so 25th term = (sum of 25 terms) - ( sum of 24 terms)
=(3×(25)²/2 +(13×25)/2) - ( 3×(24)²/2 + (13×24)/2)
= 1100 - 1020
= 80
method 2
1st term = 8
sum of 2 terms = 19
2nd term = 19 - 8= 11
so. difference = 11- 8 = 3
25th term = (8 + (25 -1)×3) = 80
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