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Question 23
If an=3−4n, then show that a1,a2,a3,⋯ form an AP. Also, find S20.
Question 23
If an=3−4n, then show that a1,a2,a3,⋯ form an AP. Also, find S20.
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Solution
Given: nth term of the series, an=3–4n
Put n=1, a1=3−4(1)=3−4=−1
Put n=2, a2=3−4(2)=3−8=−5
Put n=3, a3=3−4(3)=3−12=−9
Put n=4, a4=3−4(4)=3−16=−13
We see that,
a2−a1=−5−(−1)=−5+1=−4
a3−a2=−9−(−5)=−9+5=−4
a4−a3=−13−(−9)=−13+9=−4
a2−a1=a3−a2=a4−a3=⋯=−4
Since, the each successive term of the series has the same difference, it forms an AP.
We know that, sum of n terms of an AP,
Sn=n2[2a+(n−1)d]
∴ Sum of 20 terms of the AP,
S20=202[2(−1)+(20−1)(−4)]
=10(−2+(19)(−4))=10(−2−76)
=10×−78=−780
Hence, S20=−780
Given: nth term of the series, an=3–4n
Put n=1, a1=3−4(1)=3−4=−1
Put n=2, a2=3−4(2)=3−8=−5
Put n=3, a3=3−4(3)=3−12=−9
Put n=4, a4=3−4(4)=3−16=−13
We see that,
a2−a1=−5−(−1)=−5+1=−4
a3−a2=−9−(−5)=−9+5=−4
a4−a3=−13−(−9)=−13+9=−4
a2−a1=a3−a2=a4−a3=⋯=−4
Since, the each successive term of the series has the same difference, it forms an AP.
We know that, sum of n terms of an AP,
Sn=n2[2a+(n−1)d]
∴ Sum of 20 terms of the AP,
S20=202[2(−1)+(20−1)(−4)]
=10(−2+(19)(−4))=10(−2−76)
=10×−78=−780
Hence, S20=−780
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