Question

if the 21st term of an A.P is 25,find the sum of its 41st term

Answer 1

Answer:

25=a+(20)d  

a= 25-20d

s41=(41/2)2(25-20d)+(40d)

=(41/2)50-40d+40d

=50(41/2)

=1025

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Answer 2

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Given : a21 = 25

so,

a21= a+ 20d
25 = a+20d
a= 25- 20d


Then,

Sum of its 41st term


${s}^{41} = \frac{41}{2} (2a + 40d) \\ \\ = \frac{41}{2} (2(25 - 20d) + 40d) \\ \\ = \frac{41}{2} (2(25 - 20d) + 2(20d)) \\ \\ = \frac{41}{2} \times 2(25 - 20d + 20d) \\ \\ = 41(25) \\ \\ = 1025$


So, sum of its 41st term is 1025


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