UPSEE Solved Model Paper 6

We can solve this problem using the definition of theorem for positive integral index Binomial theorem for positive integral index If a , b are any two real numbers and n any natural number then $(a+b)^n$ = $\,{}^{n}C_{0}a^{n}b^{0}$ + $\,{}^{n}C_{1}a^{n-1}b$ + $\,{}^{n}C_{2}a^{n-2}b^{2}$ + ... + $\,{}^{n}C_{r}a^{n-r}b^{r}$ + ... + ${}^{n}C_{n}x^{0}b^{n}$ , where $\,{}^{n}C_{r}$ = $\frac{n!}{(n-r)!r!}$ Here $\,{}^{n}C_{0},\,{}^{n}C_{1},\,{}^{n}C_{2}$ ... $\,{}^{n}C_{n}$ are called binomial coefficients i.e., $(a+b)^n$ = $\sum\limits_{r=0}^{n}\,{}^{n}C_{r}a^{n-r}b^{r}$ By using this definition we can find the answer We have The coefficients of 5th, 6th and 7th terms in the expansion of $(1+x)^{}$ are in A.P. $(1+x)^n$ = $1+\,{}^{n}C_{1}x+\,{}^{n}C_{2}x^{2}+...+x^{n}$ , $\,{}^{n}C_{4},\,{}^{n}C_{5}$ and $\,{}^{n}C_{6}$ are in A.P. (since, by definition of binomial expansion) We know that If a , b , c are in A.P. , then 2b = a + c ∴ $2 \cdot \,{}^{n}C_{5}$ = $\,{}^{n}C_{4}+\,{}^{n}C_{6}$ ⇒ $2 \cdot \frac{n!}{(n-5)!5!}$ = $\frac{n!}{(n-4)!4!} + \frac{n!}{(n-6)6!}$ ⇒ $2 \cdot \frac{1}{(n-5)(n-6)!5 \times 4!}$ = $\frac{1}{(n-4)(n-5)(n-6)! \times 4!}$ + $\frac{1}{(n-6)!6 \times 5 \times 4!}$ ⇒ $\frac{2}{(n-5)5}$ = $\frac{1}{(n-4)(n-5)} + \frac{1}{30}$ ⇒ $\frac{2}{(n-5)5} - \frac{1}{(n-4)(n-5)}$ = $\frac{1}{30}$ ⇒ $\frac{2(n-4)-5}{5(n-4)(n-5)}$ = $\frac{1}{30}$ ⇒ $\frac{2(n-4)-5}{(n-4)(n-5)}$ = $\frac{5}{30}$ ⇒ $\frac{2n-8-5}{n^{2}-9n+20}$ = $\frac{5}{30}$ ⇒ $\frac{2n-13}{n^{2}-9n+20}$ = $\frac{1}{6}$ ⇒ $6(2n-13)$ = $n^{2}-9n+20$ ⇒ $n^{2} - 9n + 20$ = 12n - 78 ⇒ $n^{2}$ - 9n + 20 - 12n + 78 = 0 ⇒ $n^{2}$ - 21n + 98 = 0 ⇒ $n^{2}$ - 14n - 7n + 98 = 0 ⇒ n (n - 14) - 7 (n + 14) = 0 ⇒ (n - 7) (n - 14) = 0 ⇒ (n - 7) = 0 or (n - 14) = 0 ⇒ n = 7 or n = 14 Therefore, The coefficients of 5th, 6th and 7th terms in the expansion of $(1+x)^n$ are in A.P. then n = 7 , 14