Question

if sn=n^2+4n be the sum to first n terms of an AP ,then write the 20th term of the AP.​

Answer 1

Answer:

Sn=n^2 +4n

n=1

S1= 1^2 +4 ×1

S1= 1+4=5

S2= 2^2+4×2

S2= 4+ 8= 12

S3= 3^2+4×3

S3= 9+ 12= 21

a1= S1= 5

a2= S2-S1 = 21-12 = 9

d= a2-a1= 9-5= 4

a20= a+(n-1) d

= 5+(20-1)×4

= 5+ 19×4

= 5+ 76

= 81

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Answer 2

Answer:

we can solve this problem in atleast two ways.

we will try both here

first normal process

Sn = n^2+4n

S1 = 5 = first term

S2 = 12 = sum of two terms

second term = 12-5 = 7

common difference = 7-5 = 2

now T20 = a+(20-1)*d

= 5 + 19*2

= 5 +38

= 43

second way

S20 = 20^2 + 4*20

= 400+80 = 480

S19 = 19^2 + 4*19

= 361 + 76

= 437

T20 = S20 - S19

= 480 - 437

= 43