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CBSE Class 10 Basic Math 2026 All Sets Solved Paper

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Question : 18 of 20
Marks: +1, -0
nthn^{th} term of the A.P. −32,32,92,…-\frac{3}{2}, \frac{3}{2}, \frac{9}{2}, \dots is:
Solution:  
The first term (a) = −32-\frac{3}{2}
The common difference (d) = 32−(−32)\frac{3}{2} - \left(-\frac{3}{2}\right) = 32+32\frac{3}{2} + \frac{3}{2} = 62\frac{6}{2} = 3
The nthn^{th} term of an A.P. is given by:
an=a+(n−1)da_n = a + (n - 1)d
=−32+(n−1)3= -\frac{3}{2} + (n - 1)3
=−32+3n−3= -\frac{3}{2} + 3n - 3
=3n−(32+3)= 3n - \left(\frac{3}{2} + 3\right)
=3n−(6+32)= 3n - \left(\frac{6+3}{2}\right)
=3n−92= 3n - \frac{9}{2}
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