If 7 times the 7th term of an AP is equal to 11 times its 11th term,
Question: If 7 times the $7^{\text {th }}$ term of an AP is equal to 11 times its $11^{\text {th }}$ term, show that the $18^{\text {th }}$ term of the AP is zero. Solution: Show that: $18^{\text {th }}$ term of the AP is zero. Given: $7 a_{7}=11 a_{11}$ (Where $a_{7}$ is Seventh term, $a_{11}$ is Eleventh term, $a_{n}$ is nth term and $d$ is common difference of given AP) Formula Used: $a_{n}=a+(n-1) d$ $7(a+6 d)=11(a+10 d)$ $7 a+42 d=11 a+110 d \rightarrow 68 d=(-4 a)$ $a+17 d=0 \ldots .$ equa...
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Question: The $4^{\text {th }}$ term of an AP is three times the first and the $7^{\text {th }}$ term exceeds twice the third term by 1. Find the first term and the common difference. Solution: To Find: First term (a) and common difference (d) Given: $a_{4}=3 a_{1}$ and $a_{7}=2 a_{3}+1$ (Where $a=a_{1}$ is first term, $a_{n}$ is nth term and $d$ is common difference of given $A P$ ) Formula Used: $a_{n}=a+(n-1) d$ $a_{4}=3 a_{1} \rightarrow a+3 d=3 a \rightarrow 3 d=2 a \ldots$ equation (i) and...
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Question: If the $9^{\text {th }}$ term of an AP is 0, prove that its $29^{\text {th }}$ term is double the $19^{\text {th }}$ term. Solution: Prove that: $29^{\text {th }}$ term is double the $19^{\text {th }}$ term (i.e. $a_{29}=2 a_{19}$ ) Given: $a_{9}=0$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) Formula Used: $a_{n}=a+(n-1) d$ So $a_{9}=0 \rightarrow a+(9-1) d=0$ $a+8 d=0$ $\mathrm{a}=(-8 \mathrm{~d})$.....
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Question: The $2^{\text {nd }}, 31^{\text {st }}$ and the last terms of an AP are $7 \frac{3}{4}, \frac{1}{2}$ and $-6 \frac{1}{2}$ respectively. Find the first term and the number of terms. Solution: To Find: First term and number of terms. Given: $a_{2}=\frac{31}{4}, a_{31}=\frac{1}{2}$, and an $=\frac{-13}{2}$ Formula Used: $a_{n}=a+(n-1) d$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) By using above formula,...
Read More →The 5th and 13th terms of an AP are 5 and –3 respectively
Question: The $5^{\text {th }}$ and $13^{\text {th }}$ terms of an AP are 5 and $-3$ respectively. Find the AP and its $30^{\text {th }}$ term. Solution: To Find: AP and its $30^{\text {th }}$ term (i.e. $a_{30}=?$ ) Given: $a_{5}=5$ and $a_{13}=-3$ Formula Used: $a_{n}=a+(n-1) d$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) By using the above formula, we have $a_{5}=5=a+(5-1) d$, and $a_{13}=-3=a+(13-1) d$ $a+4...
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Question: Is $-47$ a term of the AP $5,2,-1,-4,-7, \ldots . ?$ Solution: To Find: $-47$ is a term of the AP or not. Given: The series is $5,2,-1,-4,-7, \ldots .$ $a_{1}=5, a_{2}=2$, and $d=2-5=-3($ Let suppose an $=-47)$ NOTE: n is a natural number. (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given AP) Formula Used: $a_{n}=a+(n-1) d$ $a_{n}=-47=a+(n-1) d$ $-47=5+(n-1)(-3)$ $-47-5=(n-1)(-3)$ [subtract 5 on both sides] $52=(n-1)(3)[$ ...
Read More →How many terms are there in the AP
Question: How many terms are there in the AP $1 \frac{5}{6}, 1 \frac{1}{6}, 1 \frac{1}{6}, \frac{-1}{6}, \frac{-5}{6}, \ldots,-16 \frac{1}{6} ?$ Solution: To Find: we need to find number of terms in the given AP. Given: The series is $1 \frac{5}{6}, 1 \frac{1}{6}, 1 \frac{1}{6}, \frac{-1}{6}, \frac{-5}{6}, \ldots,-16 \frac{1}{6}$. $a_{1}=1 \frac{5}{6}=\frac{11}{6}, a_{2}=1 \frac{1}{6}=\frac{7}{6}, d=\left(\frac{7}{6}\right)-\left(\frac{11}{6}\right)=\frac{-4}{6}$ and $a_{n}=-16 \frac{1}{6}=\frac...
Read More →How many terms are there in the AP
Question: How many terms are there in the AP $11,18,25,32,39, \ldots .207 ?$ Solution: To Find: we need to find a number of terms in the given AP. Given: The series is $11,18,25,32,39, \ldots 207$ $a_{1}=11, a_{2}=18, d=18-11=7$ and $a_{n}=207$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) Formula Used: $a_{n}=a+(n-1) d$ $a_{n}=207=a_{1}+(n-1)(7)$ $207-11=(n-1)(7)$ [subtract 11 on both sides] $196=(n-1)(7)$ $28=(...
Read More →Which term of the AP
Question: Which term of the AP $64,60,56,52,48, \ldots .$ is $0 ?$ Solution: To Find: we need to find $n$ when $a_{n}=0$ Given: The series is $64,60,56,52,48, \ldots$ and $a_{n}=0$ $a_{1}=64, a_{2}=60$ and $d=60-64=-4$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) Formula Used: $a_{n}=a+(n-1) d$ $a_{n}=0=a_{1}+(n-1)(-4)$ $0-64=(n-1)(-4)[$ subtract 64 on both sides $]$ $-64=(n-1)(-4)$ $64=(n-1) 4$ [Divide both sid...
Read More →Which term of the AP
Question: Which term of the AP $9,14,19,24,29, \ldots$ is $379 ?$ Solution: To Find: we need to find n when an = 379 Given: The series is $9,14,19,24,29, \ldots$ and $a_{n}=379$ $a_{1}=9, a_{2}=14$ and $d=14-9=5$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) Formula Used: $a_{n}=a+(n-1) d$ $a_{n}=379=a_{1}+(n-1) 5$ $379-9=(n-1) 5$ [subtract 9 on both side] $370=(n-1 \diamond \diamond \diamond) 5$ $74=(n-1)$ [Divi...
Read More →Find the nth term of the AP 1
Question: Find the $\mathrm{n}^{\text {th }}$ term of the AP $1, \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \ldots$ Solution: To Find: $\mathrm{n}^{\text {th }}$ term of the AP Given: The series is $1, \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \ldots$ $a_{1}=1, a_{2}=\frac{5}{6}$ and $d=\frac{5}{6}-1=\frac{-1}{6}$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given AP) Formula Used: $a_{n}=a+(n-1) d$ $a_{n}=a_{1}+(n-1)\left(\frac{-1}{6}\righ...
Read More →Find the nth term of the AP
Question: Find the $n^{\text {th }}$ term of the AP $8,3,-2,-7,-12, \ldots$ Solution: To Find: $\mathrm{n}^{\text {th }}$ term of the AP Given: The series is $8,3,-2,-7,-12, \ldots$ $a_{1}=8, a_{2}=3$ and $d=3-8=-5$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $A P$ ) Formula Used: $a_{n}=a+(n-1) d$ $a_{n}=a_{1}+(n-1)(-5)=8-(5 n-5)=8-5 n+5=13-5 n$ So the $n^{\text {th }}$ term of AP is equal to $13-5 n$...
Read More →Find the 20th term of the AP
Question: Find the $20^{\text {th }}$ term of the AP $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots .$ Solution: To Find: $20^{\text {th }}$ term of the AP Given: The series is $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots$ $a_{1}=\sqrt{2}, a_{2}=3 \sqrt{2}$ and $d=3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $A P$ ) Formula Used: $a_{n}=a+(n-1) d$ $a_{20}=a_{1}+(20-1)(2 \sqrt{2})=...
Read More →Find the 23rd term of the AP 7, 3, 1, –1, –3,
Question: Find the $23^{\text {rd }}$ term of the AP $7,3,1,-1,-3, \ldots$ Solution: To Find: $23^{\text {rd }}$ term of the AP Given: The series is $7,5,3,1,-1,-3, \ldots$ $a_{1}=7, a_{2}=5$ and $d=3-5=-2$ (Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $A P$ ) Formula Used: $a_{n}=a+(n-1) d$ So put n =23 in above formula, we have $a_{23}=a_{1}+(23-1)(-2)=7-44=-37$ So $23^{\text {rd }}$ term of AP is equal to $-37$....
Read More →Find the first 5 terms of the sequence, defined by
Question: Find the first 5 terms of the sequence, defined by $a_{1}=-1, a_{n}=$ $\frac{a_{n-1}}{n}$ for $n \geq 2$ Solution: To Find: First five terms of a given sequence Condition: n 2 $a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}$ for $n \geq 2$ Put $n=2$ in $n^{\text {th }}$ term (i.e. $a_{n}$ ), we have $a_{2}=\frac{(-1)}{2}\left(\right.$ as $\left.a_{1}=-1\right)$ Put $n=3$ in $n^{\text {th }}$ term $\left(\right.$ i.e. $\left.a_{n}\right)$, we have $a_{3}=\frac{(-1)}{6}\left(\operatorname{as} a_{2}=\...
Read More →Find the first five terms of the sequence, defined by
Question: Find the first five terms of the sequence, defined by $a_{1}=1, a_{n}=a_{n-1}+3$ for $n \geq 2$ Solution: To Find: First five terms of a given sequence. Condition: $\mathrm{n} \geq 2$ Given: $a_{1}=1, a_{n}=a_{n-1}+3$ for $n \geq 2$ Put $n=2$ in $n^{\text {th }}$ term $\left(\right.$ i.e. $\left.a_{n}\right)$, we have $a_{2}=a_{2-1}+3=a_{1}+3=1+3=4\left(\right.$ as $\left.a_{1}=1\right)$ Put $\mathrm{n}=3$ in $\mathrm{n}^{\text {th }}$ term (i.e. $\mathrm{a}_{\mathrm{n}}$ ), we have $a...
Read More →Write first 4 terms in each of the sequences:
Question: Write first 4 terms in each of the sequences: (i) $a_{n}=(5 n+2)$ (ii) $a_{n}=\frac{(2 n-3)}{4}$ (iii) $a_{n}=(-1)^{n-1} \times 2^{n+1}$ Solution: To Find: First four terms of given series. (i) Given: $\mathrm{n}^{\text {th }}$ term of series is $(5 \mathrm{n}+2)$ Put $n=1,2,3,4$ in $n^{\text {th }}$ term, we get first (a1), Second (a2), Third (a3) \ Fourth (a4) term $a_{1}=(5 \times 1+2)=7$ $a_{2}=(5 \times 2+2)=12$ $a_{3}=(5 \times 3+2)=17$ $a_{4}=(5 \times 4+2)=22$ First four terms ...
Read More →Find the positive value of m for which the coefficient of
Question: Find the positive value of $m$ for which the coefficient of $x^{2}$ in the expansion of $(1+x)^{m}$ is 6 Solution: To find: the positive value of m for which the coefficient of x2 in the expansion of $(1+x)^{m}$ is 6 Formula Used: General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expression, $(1+x)^{m}$, we get $\mathrm{...
Read More →In the binomial expansion of
Question: In the binomial expansion of $(a+b)^{n}$, the coefficients of the $4^{\text {th }}$ and $13^{\text {th }}$ terms are equal to each other. Find the value of $\mathrm{n}$. Solution: To find: the value of n with respect to the binomial expansion of (a + b)n where the coefficients of the 4th and 13th terms are equal to each other Formula Used: A general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{\mathrm{n}} \mathrm{C}_{\...
Read More →Find the coefficient of
Question: Find the coefficient of $x^{n}$ in the expansion of $(1+x)(1-x)^{n}$. Solution: To find: the coefficient of $x^{n}$ in the expansion of $(1+x)(1-x)^{n}$. Formula Used: Binomial expansion of $(x+y)^{n}$ is given by, $(\mathrm{x}+\mathrm{y})^{\mathrm{n}}=\sum_{r=0}^{n}\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}$ Thus, $(1+x)(1-x)^{n}$ $=(1+x)\left(\left(\begin{array}{l}n \\ 0\end{array}\right)(-x)+\lef...
Read More →Write the 4th term from the end in the expansion of
Question: Write the $4^{\text {th }}$ term from the end in the expansion of $\left(\frac{3}{x^{2}}-\frac{x^{3}}{6}\right)^{7}$ Solution: To find: $4^{\text {th }}$ term from the end in the expansion of $\left(\frac{3}{x^{2}}-\frac{x^{3}}{6}\right)^{7}$ Formula Used: A general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{\mathrm{n}} \mathrm{C}_{r}=\frac{n !}{r !(n-r) !}$ Total number of terms in the expansion is 8 Thus, the $4^{\...
Read More →If the coefficients of
Question: If the coefficients of $(r-5)$ th and $(2 r-1)$ th terms in the expansion of $(1+$$x)^{34}$ are equal, find the value of r. Solution: To find: the value of r with respect to the binomial expansion of $(1+$$x)^{34}$ where the coefficients of the (r 5)th and (2r 1)th terms are equal to each other Formula Used: The general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ Now, finding the (r 5)...
Read More →Write the coefficient of
Question: Write the coefficient of $x^{7} y^{2}$ in the expansion of $(x+2 y)^{9}$ Solution: To find: the coefficient of $x^{7} y^{2}$ in the expansion of $(x+2 y)^{9}$ Formula Used: A general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expression, $(x+2 y)^{9}$, we get $T_{r+1}={ }^{9} C_{r} \times x^{9-r} \times(2 y)^{r}$ The value of $r$ for which coeffici...
Read More →Which term is independent of x in the expansion of
Question: Which term is independent of x in the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9} ?$ Solution: To find: the term independent of x in the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9} ?$ Formula Used: A general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expression, $\left(x-\frac{1}{3 x^{2}}\right)^{9}$ we get $T_{r+1}={ }^{9} C_{r} \tim...
Read More →Which term is independent of x in the expansion of
Question: Which term is independent of x in the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9} ?$ Solution: To find: the term independent of x in the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9} ?$ Formula Used: A general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by, $T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ Now, finding the general term of the expression, $\left(x-\frac{1}{3 x^{2}}\right)^{9}$ we get $T_{r+1}={ }^{9} C_{r} \tim...
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