A polygon has 25 sides, the length of which, starting from the smallest side are in AP. If the perimeter of the polygon is 1100 cm and the length of the largest side is 10 times that of the smallest, find the length of the smallest and the common difference of the AP.
Answers
Answered by
0
Answer:
Let a be the length of smallest side and d cm the common difference ltBrgt Now
S_(n)=(n)/(2)[2a+(n-1)d]
<br> n = 25,
S_(25)= 2100
ltBrgt
2100=(25)/(2)[2a+24d]" "...(1)
<br> a+12d = 84 <br>
because
The largest side =
25^(th)
side ltBrgt =a+24d = 20a
" "...(2)
<br> solve (1) and (2) <br> a = 8 , d =
6(1)/(3)
Answered by
1
Answer:
Let a be the length of smallest side and d cm the common difference ltBrgt Now
S_(n)=(n)/(2)[2a+(n-1)d]
<br> n = 25,
S_(25)= 2100
ltBrgt
2100=(25)/(2)[2a+24d]" "...(1)
<br> a+12d = 84 <br>
because
The largest side =
25^(th)
side ltBrgt =a+24d = 20a
" "...(2)
<br> solve (1) and (2) <br> a = 8 , d =
6(1)/(3)
Step-by-step explanation:
Hope it is helpful
Please Mark me as brainliest
Similar questions