Question

in an a.p. the sum of first 41 terms 5125 to find 21st term of the a.p.​

Answer 1

here your answer ..hope u pasand

Answer 2

Step-by-step explanation:

Tn=a+(n−1)d

50=a+(21−1)d

50=a+20d...(1)

S

n

=

2

n

[2a+(n−1)d]

S

41

=

2

41

[2a+(41−1)d]

=

2

41

[2a+40d]

=

2

41

2[a+20d]

=41[a+20d]

From (1)

=41[50]

S

41

=2050

mark brainliest