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Solution

✅ Correct Option: 2

The 3rd term of an A.P. is 1 and the 9th term is 19.

In an Arithmetic Progression, the nth term is given by:

an=a+(n−1)da_n = a + (n - 1)d

where aa is the first term and dd is the common difference.


For the 3rd term:

a3=a+2da_3 = a + 2d

a+2d=1a + 2d = 1 ... (Equation 1)

For the 9th term:

a9=a+8da_9 = a + 8d

a+8d=19a + 8d = 19 ... (Equation 2)


Subtracting Equation 1 from Equation 2:

(a+8d)−(a+2d)=19−1(a + 8d) - (a + 2d) = 19 - 1

6d=186d = 18

d=3d = 3


Substituting d=3d = 3 into Equation 1:

a+2(3)=1a + 2(3) = 1

a+6=1a + 6 = 1

a=−5a = -5


The 23rd term is:

a23=a+22da_{23} = a + 22d

a23=−5+22(3)a_{23} = -5 + 22(3)

a23=−5+66a_{23} = -5 + 66

a23=61a_{23} = 61

Therefore, the 23rd term is 61.

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