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Suppose x1,x2,x3,…,x100x_{1}, x_{2}, x_{3}, \ldots, x_{100} are in arithmetic progression such that x5=−4x_{5}=-4 and 2x6+2x9=x11+x132 x_{6}+2 x_{9}=x_{11}+x_{13}, Then, x100x_{100} equals

Solution

✅ Correct Option: 2

The formula for the nth term of an AP is: xn=a+(n−1)dx_n = a + (n-1)d


x5=a+(5−1)d=a+4dx_5 = a + (5-1)d = a + 4d

Since x5=−4x_5 = -4:

a+4d=−4a + 4d = -4 ... (equation 1)


Given: 2x6+2x9=x11+x132x_6 + 2x_9 = x_{11} + x_{13}

2(x6+x9)=x11+x132(x_6 + x_9) = x_{11} + x_{13}

x6=a+5dx_6 = a + 5d

x9=a+8dx_9 = a + 8d

x11=a+10dx_{11} = a + 10d

x13=a+12dx_{13} = a + 12d

Substituting:

x6+x9=(a+5d)+(a+8d)=2a+13dx_6 + x_9 = (a + 5d) + (a + 8d) = 2a + 13d

x11+x13=(a+10d)+(a+12d)=2a+22dx_{11} + x_{13} = (a + 10d) + (a + 12d) = 2a + 22d

Our equation becomes:

2(2a+13d)=2a+22d2(2a + 13d) = 2a + 22d

4a+26d=2a+22d4a + 26d = 2a + 22d


From our simplified equation:

4a+26d=2a+22d4a + 26d = 2a + 22d

4a−2a=22d−26d4a - 2a = 22d - 26d

2a=−4d2a = -4d

a+2d=0a + 2d = 0 ... (equation 2)

Now we have two equations:

Equation 1: a+4d=−4a + 4d = -4

Equation 2: a+2d=0a + 2d = 0

Subtracting equation 2 from equation 1:

(a+4d)−(a+2d)=−4−0(a + 4d) - (a + 2d) = -4 - 0

a+4d−a−2d=−4a + 4d - a - 2d = -4

2d=−42d = -4

d=−2d = -2


Finding aa: Substitute d=−2d = -2 into equation 2:

a+2(−2)=0a + 2(-2) = 0

a−4=0a - 4 = 0

a=4a = 4


Using our AP formula with a=4a = 4 and d=−2d = -2:

x100=a+(100−1)dx_{100} = a + (100-1)d

=4+99(−2) = 4 + 99(-2)

=4−198 = 4 - 198

=−194 = -194

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