The consecutive terms whose coefficients are equal in the expansion of $(8 + 7 x)^{44}$ are

t_{21}, t_{22}

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t_{23}, t_{24}

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t_{22}, t_{23}

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t_{20}, t_{21}

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Solution

The correct option is A $t_{21}, t_{22}$
We know, $T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$
Applying to the above question, we get
$^{44} C_{r} 8^{44 - r} 7^{r} =^{44} C_{r + 1} 8^{44 - r - 1} 7^{r + 1}$
$\frac{44! .8}{(44 - r)! (r)!} = d \frac{44! .7}{(44 - r - 1)! (r + 1)!}$
$\frac{8}{44 - r} = \frac{7}{r + 1}$
$8 r + 8 = 308 - 7 r$
$15 r = 300$
$∴ r = 20$
Hence, the terms are $T_{r + 1}$ and $T_{r + 2}$ .
Substituting the value of $r$ , we get the consecutive terms as $t_{21}$ , $t_{22}$ .

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