If the 21 th term of an A. P.is 25. Find the sum of its first 41 terms
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Given:- a21=25
since, an=a+(n-1)d
=) a21=a+20d ------1)
since , a21=25 (given)
=) a+20d=25
=)a=25-20d -------2)
Since, sum of 'x' terms of an A.P=) Sn=n/2[2a+(n-1)d]
=)41st term (s41) =) 41/2(2(25 - 20d) + (41 - 40) × d
= 41/2(2(25 - 20d) + 40d)
= 41/2(50 - 40d + 40d)
= 41/2(50)
= 41 × 25
= 1025.
Therefore, the sum( Sn) = 1025..
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since, an=a+(n-1)d
=) a21=a+20d ------1)
since , a21=25 (given)
=) a+20d=25
=)a=25-20d -------2)
Since, sum of 'x' terms of an A.P=) Sn=n/2[2a+(n-1)d]
=)41st term (s41) =) 41/2(2(25 - 20d) + (41 - 40) × d
= 41/2(2(25 - 20d) + 40d)
= 41/2(50 - 40d + 40d)
= 41/2(50)
= 41 × 25
= 1025.
Therefore, the sum( Sn) = 1025..
Hope it Helps ..
Mark as Brainliest
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