ICSE BoardEnglish MediumSTD 10MathematicsArithmetic Progression3 Marks
Question
Can $2 n^2-7$ be the $n^{\text {th }}$ term of an $A.P?$ Explain.
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Answer
We have $2 n^2-7$
Substitute $n =1,2,3, \ldots$, we get
$2(1)^2-7,2(2)^2-7,2(3)^2-7,2(4)^2-7, \ldots$
$-5,1,11, \ldots$
Difference between the first and second term $=1-(-5)=6$
And Difference between the second and third term $=11-1=10$
Here, the common difference is not same.
Therefore the $n ^{\text {th }}$ term of an A.P can't be $2 n^2-7$.
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