Question

if the 21st term of an AP is 25 ,find the sum of its first 41 terms​

Answer 1

Answer: 1025

Step-by-step explanation:

We know, from the equation for a term of an A.P.:

$a+(20)d=25$

We need to find :

$\frac{41}{2} (2a+40d)$

We see:

$2a+40d=2(a+20d)=50$

∴Answer is $(41/2)*50=1025$

Answer 2

Answer:

tn=a+(n-1)d

t25=a+(21-1)d

25=a+20d eq. 1

we know that Sn =n/2[2a+(n-1)d]

S41=n/2[2a+40d]

From eq. 1

a+20d=25

2[a+20d]=50

Then s41=41/2*50

S41=41*25=1025