1f:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 20:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1e:T43f,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"Let the coefficients of three consecutive terms in the binom","item":"https://www.divinejee.com/question/let-the-coefficients-of-three-consecutive-terms-in-the-binomial-expansion-of-1-2xn-be-in-the-ratio-2"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question Let the coefficients of three consecutive terms in the binom on Topic","text":"

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio 2 : 5 : 8. Then the coefficient of the term...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/let-the-coefficients-of-three-consecutive-terms-in-the-binomial-expansion-of-1-2xn-be-in-the-ratio-2"}}}]}21:T508,$\mathrm{t}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2 \mathrm{x})^{\mathrm{r}}$

$$ \begin{aligned} & \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\ & \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\ & \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{r}+1}}=\frac{5}{8} \\\\ & \Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \\\\ & \Rightarrow 4 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2) \end{aligned} $$

From (1) and (2)

$$ \Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8 $$

$(1) \Rightarrow r=4$

so, coefficient of middle term is

$$ { }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120 $$