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Answer:
(I) Given:
AP₁: a=8, d=20
AP₂: a=–30, d=8
Here, (AP₁) Sn=S₂n (AP₂)
To Find: Value of 'n'
∵, Sn=n/2[2a+(n–1)d]
So, For AP₁:
Sn=n/2[2(8)+(n–1)(20)]
Sn=n/2[16+20n–20]
=n/2[20n–4]
Sn=n[10n–2]
For AP₂:
S₂n=2n/2[2(–30)+(2n–1)(8)]
=n[–60+16n–8]
S₂n=n[16n–68]
Now, ∵; Sn=S₂n (given)
So, n[10n–2]=n[16n–68]
16n–10n=68–2
6n=66
n=11
(II) As the total number of flags is 27, so there would be 13 flags on one side, 13 flags on the other side and one flag in the middle.
Consider that she fixes 13 flags on one side of the middle flag.
So, the distance covered by Ruchi to fix the first flag and return is 2m+2m=4m.
The distance covered by Ruchi to fix the second flag and return is 4m+4m=8m.
Similarly, the distance covered by Ruchi to fix the thirteenth flag and return is
4m+8m+12m⋯+13terms.
It will form an AP;
AP: 4m+8m+12m⋯+13terms
∵, Sn=n/2[2a+(n–1)d]
Sn= 13/2 [2(4)+(13−1)4]
= 13/2 (56)
=364m
The total distance travelled by Ruchi to fix the flags and return back is 2×364=728m.
The total distance covered by Ruchi only to carry the flag is
=>1/2 ×728=
364m
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