If the coefficients of 3 consecutive terms in | vedclass

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(D) Let the three consecutive terms be $T_{r+1}, T_{r+2}, T_{r+3}$. Their coefficients are $^{23}C_r, ^{23}C_{r+1}, ^{23}C_{r+2}$.Given that these coe

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$T_{14}, T_{15}, T_{16}$

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(D) Let the three consecutive terms be $T_{r+1}, T_{r+2}, T_{r+3}$. Their coefficients are $^{23}C_r, ^{23}C_{r+1}, ^{23}C_{r+2}$.Given that these coefficients are in arithmetic progression,we have:$2 \cdot ^{23}C_{r+1} = ^{23}C_r + ^{23}C_{r+2}$Dividing by $^{23}C_{r+1}$,we get:$2 = \frac{^{23}C_r}{^{23}C_{r+1}} + \frac{^{23}C_{r+2}}{^{23}C_{r+1}}$Using the formula $\frac{^{n}C_{k-1}}{^{n}C_k} = \frac{k}{n-k+1}$,we have:$2 = \frac{r+1}{23-r} + \frac{23-(r+1)+1}{r+2} = \frac{r+1}{23-r} + \frac{23-r}{r+2}$Let $x = 23-r$. Then $r+1 = 24-x$ and $r+2 = 25-x$.$2 = \frac{24-x}{x} + \frac{x}{25-x}$$2x(25-x) = (24-x)(25-x) + x^2$$50x - 2x^2 = 600 - 49x + x^2 + x^2$$4x^2 - 99x + 600 = 0$Solving this quadratic equation: $(x-25)(4x-24) = 0$ is not correct. Let's re-evaluate: $4x^2 - 99x + 600 = 0$ gives $x=15$ or $x=10$.If $x=15$,$23-r=15 \Rightarrow r=8$. The terms are $T_9, T_{10}, T_{11}$.If $x=10$,$23-r=10 \Rightarrow r=13$. The terms are $T_{14}, T_{15}, T_{16}$.Thus,the terms are $T_{14}, T_{15}, T_{16}$.

If the coefficients of $3$ consecutive terms in the expansion of $(1+x)^{23}$ are in arithmetic progression,then those terms are

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