InterviewSolution
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InterviewSolution
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If `a_(n)=3-4n`, then show that `a_(1),a_(2),a_(3), …` form an AP. Also, find `S_(20)`. |
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Answer» Given that, `n`th term of the series is `a_(n)=3-4n " " `…(i) Put`n=1, " " a_(1)=3-4(1)=3-4= -1` Put`n=2, " " a_(2)=3-4(2)= 3-8=-5` Put`n=3, " " a_(3)=3-4(3)= 3-12=-9` Put`n=4, " " a_(4)=3-4(4)=3-16= -13` So the series becomes `-1,-5,-9,-13, …` We see that, `a_(2)-a_(1)=-5-(-1)=-5+1=-4` `a_(3)-a_(2)=-9-(-5)=-9+5= -4` `a_(4)-a_(3)= -13-(-9)=-13+9= -4` ` "i.e., " a_(2)-a_(1)=a_(3)-a_(2)=a_(4)-a_(3)=...= -4` Since, the each successive term of the series has the same difference. So, it forms an AP. We know that, sun of n terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d]` ` :. ` Sum of 20 terms of the AP,`S_(20)=(20)/(2)[2(-1)+(20-1)(-4)]` `S_(20)=10(-2+(19)(-4))=10(-2-76)` ` " " =10xx-78= -780` Hence, the required sum of 20 terms i.e., `S_(20)` is `-780`. |
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