Question 11 Mark
Find the sum to n terms in each of the series in whose $n^{th}$ terms is given by $n^2 + 2^n$.
AnswerGiven: $a_n = n^2 + 2^n$
$\therefore {S_n} = \sum\limits_{k = 1}^n {{a_k}} =\sum\limits_{ k = 1}^n {{k^2}} + {2^k}$
$=\left(1^2+2^1\right)+\left(2^2+2^2\right)+\left(3^2+2^3\right) \ldots \ldots\left(n^2+2^n\right)$
$=\left(1^2+2^3+3^2+\ldots . .+n^2\right)+\left(2^1+2^2+2^3+\ldots \ldots+2^n\right)$
= $\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { 2 \left( 2 ^ { n } - 1 \right) } { 2 - 1 }$
= $\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + 2 \left( 2 ^ { n } - 1 \right)$
View full question & answer→Question 21 Mark
Find the sum to n terms of the series $\frac { 1 } { 1 \times 2 } + \frac { 1 } { 2 \times 3 } + \frac { 1 } { 3 \times 4 } + \ldots$
AnswerLet the given series be S = $\frac { 1 } { 1 \times 2 } + \frac { 1 } { 2 \times 3 } + \frac { 1 } { 3 \times 4 } + \ldots$
Then, $n^{th}$ term $T_n$ = $\frac { 1 } { n ( n + 1 ) }$
Now, we will split the denominator of the $n^{th}$ term into two parts or we will write $T_n$ as the difference of two terms.
$\therefore T_n = \frac { 1 } { n ( n + 1 ) } = \frac { ( n + 1 ) - n } { n ( n + 1 ) }$
= $\frac { n + 1 } { n ( n + 1 ) } - \frac { n } { n ( n + 1 ) } = \frac { 1 } { n } - \frac { 1 } { n + 1 }$
On putting n = 1, 2, 3, 4,... successively, we get
$T_1 = \frac { 1 } { 1 } - \frac { 1 } { 2 }$
$T_2 = \frac { 1 } { 2 } - \frac { 1 } { 3 }$
$T_3 = \frac { 1 } { 3 } - \frac { 1 } { 4 }$................
$T_n = \frac { 1 } { n } - \frac { 1 } { n + 1 }$
On adding all these terms, we get
$S = T_1 + T_2 + T_3 + ...+ T_n$
= $\left( \frac { 1 } { 1 } - \frac { 1 } { 2 } \right) + \left( \frac { 1 } { 2 } - \frac { 1 } { 3 } \right) + \ldots + \left( \frac { 1 } { n } + \frac { 1 } { n + 1 } \right)$
= 1 - $\frac { 1 } { n + 1 } = \frac { 1 } { 1 } - \frac { 1 } { n + 1 } = \frac { n + 1 - 1 } { n + 1 }$
$\Rightarrow$ S = $\frac { n } { n + 1 }$
View full question & answer→Question 31 Mark
Find the indicated terms of the sequence, whose nth term is $a _ { n } = ( - 1 ) ^ { n - 1 } n ^ { 3 }; a_9$
AnswerGiven: $a_n=(-1)^{n-1} n^3$
$\therefore a_9=(-1)^{9-1} \times(9)^3=(-1)^8 \times 729=729$
Therefore, $9^{\text {th }}$ term is 729 .
View full question & answer→Question 41 Mark
Find the indicated terms of the sequence, whose nth term is $a _ { n } = \frac { n ^ { 2 } } { 2 ^ { n } }; a_7$
AnswerGiven: $a _ { n } = \frac { n ^ { 2 } } { 2 ^ { n } }$
$\therefore a _ { 7 } = \frac { 7 ^ { 2 } } { 2 ^ { 7 } } = \frac { 49 } { 128 }$
Therefore, $7^{th}$ term is $\frac { 49 } { 128 }.$
View full question & answer→Question 51 Mark
Find the indicated terms of the sequence, whose $n$th term is $a_n=4 n-3 ; a_{17}, a_{24}$.
AnswerWe have, $a_n=4 n-3$
On putting $n=17$, we get
$a_{17}=4 \times 17-3=68-3=65$
On putting $n=24$, we get
$a_{24}=4 \times 24-3=96-3=93$
View full question & answer→Question 61 Mark
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = n . \frac { n ^ { 2 } + 5 } { 4 }$ .
AnswerGiven: $a_{n}=n \cdot \frac{n^{2}+5}{4}$
Putting n = 1,2,3,4 and 5, we get,
$a_{1}=1 \frac{1^{2}+5}{4}$$=1 . \frac{1+5}{4}=\frac{6}{4}=\frac{3}{2}$
$a_{2}=2 \cdot \frac{2^{2}+5}{4}$$=2 . \frac{4+5}{4}=\frac{18}{4}=\frac{9}{2}$
$a_{3}=3 . \frac{3^{2}+5}{4}$$=3 . \frac{9+5}{4}=3 \times \frac{14}{4}$$=\frac{42}{4}=\frac{21}{2}$
$a_{4}=4 \cdot \frac{4^{2}+5}{4}$$=4 . \frac{16+5}{4}=\frac{84}{4}$ = 21
$a_{5}=5 \cdot \frac{5^{2}+5}{4}$$=5 . \frac{25+5}{4}=5 \times \frac{30}{4}$$=\frac{150}{4}=\frac{75}{2}$
Therefore, the first five terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}$, 21 and $\frac{75}{2}$
View full question & answer→Question 71 Mark
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = ( - 1 ) ^ { n - 1 } \cdot 5 ^ { n + 1 }$
AnswerGiven: $a _ { n } = ( - 1 ) ^ { n - 1 } \cdot 5 ^ { n + 1 }$Putting n = 1, 2, 3, 4 and 5, we get,
$a _ { 1 } = ( - 1 ) ^ { 1 - 1 } \cdot 5 ^ { 1 + 1 } = ( - 1 ) ^ { 0 } \cdot 5 ^ { 2 } = 1 \times 25 = 25$
$a _ { 2 } = ( - 1 ) ^ { 2 - 1 } 5 ^ { 2 + 1 } = ( - 1 ) ^ { 1 } \cdot 5 ^ { 3 } = - 1 \times 125 = - 125$
$a _ { 3 } = ( - 1 ) ^ { 3 - 1 } \cdot 5 ^ { 3 + 1 } = ( - 1 ) ^ { 2 } \cdot 5 ^ { 4 } = 1 \times 625 = 625$
$a _ { 4 } = ( - 1 ) ^ { 4 - 1 } \cdot 5 ^ { 4 + 1 } = ( - 1 ) ^ { 3 } \cdot 5 ^ { 5 } = - 1 \times 3125 = - 3125$
$a _ { 5 } = ( - 1 ) ^ { 5 - 1 } \cdot 5 ^ { 5 + 1 } = ( - 1 ) ^ { 4 } \cdot 5 ^ { 6 } = 1 \times 15625 = 15625$
Therefore, the first five terms are 25, -125, 625, -3125 and 15625.
View full question & answer→Question 81 Mark
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = \frac { 2 n - 3 } { 6 }$.
AnswerGiven: $a _ { n } = \frac { 2 n - 3 } { 6 }$Putting n = 1, 2, 3, 4 and 5, we get,
$a _ { 1 } = \frac { 2 \times 1 - 3 } { 6 } = \frac { 2 - 3 } { 6 } = \frac { - 1 } { 6 }$
$a _ { 2 } = \frac { 2 \times 2 - 3 } { 6 } = \frac { 4 - 3 } { 6 } = \frac { 1 } { 6 }$
$a _ { 3 } = \frac { 2 \times 3 - 3 } { 6 } = \frac { 6 - 3 } { 6 } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
$a _ { 4 } = \frac { 2 \times 4 - 3 } { 6 } = \frac { 8 - 3 } { 6 } = \frac { 5 } { 6 }$
$a _ { 5 } = \frac { 2 \times 5 - 3 } { 6 } = \frac { 10 - 3 } { 6 } = \frac { 7 } { 6 }$
Therefore, the first five terms are $\frac { - 1 } { 6 } , \frac { 1 } { 6 } , \frac { 1 } { 2 } , \frac { 5 } { 6 }$ and $\frac { 7 } { 6 }.$
View full question & answer→Question 91 Mark
Write the first five terms of the sequence whose $n^{th}$ term is $a_n = 2^n$
AnswerGiven: $a_n=2^n$ Putting $n=1,2,3,4$ and 5 , we get,
$a_1=2^1=2$
$a_2=2^2=4$
$a_3=2^3=8$
$a_4=2^4=16$
$a_5=2^5=32$
Therefore, the first five terms are $2,4,8,16$ and 32 .
View full question & answer→Question 101 Mark
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = \frac { n } { n + 1 }$
AnswerGiven: $a _ { n } = \frac { n } { n + 1 }$Putting n = 1, 2, 3, 4 and 5, we get,
$a _ { 1 } = \frac { 1 } { 1 + 1 } = \frac { 1 } { 2 }$
$a _ { 2 } = \frac { 2 } { 2 + 1 } = \frac { 2 } { 3 }$
$a _ { 3 } = \frac { 3 } { 3 + 1 } = \frac { 3 } { 4 }$
$a _ { 4 } = \frac { 4 } { 4 + 1 } = \frac { 4 } { 5 }$
$a _ { 5 } = \frac { 5 } { 5 + 1 } = \frac { 5 } { 6 }$
Therefore, the first five terms are $\frac { 1 } { 2 } , \frac { 2 } { 3 } , \frac { 3 } { 4 } , \frac { 4 } { 5 }$ and $\frac { 5 } { 6 }.$
View full question & answer→Question 111 Mark
The Fibonacci sequence is defined by $1 = a_1 = a_2$ and $a_n = a_{n-1}+ a_{n-2}, n > 2$. Find $\frac { a _ { n + 1 } } { a _ { n } }$, for $n = 1, 2, 3, 4, 5$.
AnswerGiven, $1=a_1=a_2$
and $a_n=a_{n-1}+a_{n-2}, n>2$
On putting $n=3,4,5,6$ respectively, we get
For $n=3, a_3=a_{3-1}+a_{3-2}=a_2+a_1=1+1=2$
For $n=4, a_4=a_{4-1}+a_{4-2}=a_3+a_2=2+1=3$
For $\mathrm{n}=5, \mathrm{a}_5=\mathrm{a}_{5-1}+\mathrm{a}_{5-2}=\mathrm{a}_4+\mathrm{a}_3=3+2=5$
For $n=6, a_6=a_{6-1}+a_{6-2}=a_5+a_4=5+3=8$
Now, $\frac{a_{n+1}}{a_n}$, for $n=1,2,3,4,5$.
For n = 1, $\frac { a _ { 2 } } { a _ { 1 } } = \frac { 1 } { 1 }$ = 1
For n = 2, $\frac { a _ { 3 } } { a _ { 2 } } = \frac { 2 } { 1 }$ = 2
For n = 3, $\frac { a _ { 4 } } { a _ { 3 } } = \frac { 3 } { 2 }$
For n = 4, $\frac { a _ { 5 } } { a _ { 4 } } = \frac { 5 } { 3 }$
For n = 5, $\frac { a _ { 6 } } { a _ { 5 } } = \frac { 8 } { 5 }$
Hence, the terms are 1, 2, $\frac { 3 } { 2 } , \frac { 5 } { 3 }$ and $\frac { 8 } { 5 }$
View full question & answer→Question 121 Mark
Write the first five terms of the sequence and obtain the corresponding series $a_1 = a_2 = 2, a_n = a_{n-1} - 1, n > 2$.
AnswerGiven: $a_1=a_2=2, a_n=a_{n-1}-1, n>2$
Putting $n=3,4$ and 5 , we get
$a_3=a_{3-1}-1=a_{2-1}=2-1=1$
$a_4=a_{4-1}-1=a_{3-1}=1-1=0$
$a_5=a_{5-1}-1=a_{4-1}=0-1=-1$
Hence the first five terms are $2,2,1,0,-1$.
Therefore, corresponding series is $2+2+1+0+(-1)+\ldots . .$.
View full question & answer→Question 131 Mark
Write the first five terms of the sequence and obtain the corresponding series $a_1 = -1, a _ { n } = \frac { a _ { n - 1 } } { n } , n \geq 2$
AnswerGiven: $a_1 = -1, a _ { n } = \frac { a _ { n - 1 } } { n } , n \geq 2$Putting n = 2, 3, 4 and 5, we get
$a _ { 2 } = \frac { a _ { 2 - 1 } } { 2 } = \frac { a _ { 1 } } { 2 } = \frac { - 1 } { 2 }$
$a _ { 3 } = \frac { a _ { 3 - 1 } } { 3 } = \frac { a _ { 2 } } { 3 } = \frac { - 1 / 2 } { 3 } = \frac { - 1 } { 6 }$
$a _ { 4 } = \frac { a _ { 4 - 1 } } { 4 } = \frac { a _ { 3 } } { 4 } = \frac { - 1 / 6 } { 4 } = \frac { - 1 } { 24 }$
$a _ { 5 } = \frac { a _ { 5 - 1 } } { 5 } = \frac { a _ { 4 } } { 5 } = \frac { - 1 / 24 } { 5 } = \frac { - 1 } { 120 }$
Hence the first five terms are $- 1,{{ - 1} \over 2},{{ - 1} \over 6},{{ - 1} \over {24}},{{ - 1} \over {120}}$
$\therefore$ Corresponding series is $- 1 + \left( {{{ - 1} \over 2}} \right) + \left( {{{ - 1} \over 6}} \right) + \left( {{{ - 1} \over {24}}} \right) + \left( {{{ - 1} \over {120}}} \right).........$
View full question & answer→Question 141 Mark
Write the first five terms of the sequence and obtain the corresponding series $a_1 = 3, a_n = 3a_{n-1} + 2$ for all $n > 1$
AnswerGiven: $a_1=3, a_{n-1}+2$ for all $n>1$ Putting $n=2,3,4$ and 5 , we get
$a_2=3 a_{2-1}+2=3 a_{2-1}+2=3 a_1+2=3 \times 3+2=9+2=11$
$a_3=3 a_{3-1}+2=3 a_2+2=3 \times 11+2=33+2=35$
$a_4=3 a_{4-1}+2=3 a_3+2=3 \times 35+2=105+2=107$
$a_5 3 a_{5-1}+2=3 a_4+2=3 \times 107+2=321+2=323$
Hence the first five terms are $3,11,35,107,323$.
Therefore, corresponding series is $3+11+35+107+323+\ldots \ldots . . .$.
View full question & answer→Question 151 Mark
Find the indicated terms of the sequence, whose nth term is $a _ { n } = \frac { n ( n - 2 ) } { n + 3 }; a_{20}$
AnswerGiven: $a_{n}=\frac{n(n-2)}{n+3}$
$\therefore a_{20}=\frac{20(20-2)}{20+3}$$=\frac{20 \times 18}{23}=\frac{360}{23}$
Therefore, $20^{th}$ term is $\frac{360}{23}$.
View full question & answer→Question 161 Mark
Write the first five terms of the sequence whose $n^{th}$ term is $a_n = n (n + 2)$
AnswerGiven: $a_n = n(n + 2)$
Putting n = 1, 2, 3, 4 and 5, we get,
$a _ { 1 } = 1 ( 1 + 2 ) = 1 \times 3 = 3$
$a _ { 2 } = 2 ( 2 + 2 ) = 2 \times 4 = 8$
$a _ { 3 } = 3 ( 3 + 2 ) = 3 \times 5 = 15$
$a _ { 4 } = 4 ( 4 + 2 ) = 4 \times 6 = 24$
$a _ { 5 } = 5 ( 5 + 2 ) = 5 \times 7 = 35$
Therefore, the first five terms are 3, 8, 15, 24 and 35.
View full question & answer→Question 171 Mark
Find the 10th and nth terms of the G.P. 5, 25,125,….
AnswerWe have, $5+25+125+\ldots$ is GP.
Here, $a=5$ and $r=\frac{25}{5}=5$
We know that, $T_n=a r^{n-1}=5(5)^{n-1}=5^n$ and $T_{10}=5(5)^{10-1}=5^{10}$
View full question & answer→Question 181 Mark
Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P.
AnswerLet $A_1, A_2, A_3, A_4, A_5$ and $A_6$ be six numbers between 3 and 24 such that $3, A_1, A_2, A_3, A_4, A_5, A_6, 24$ are in $A . P$. Here, a $=3, b=24, n=8$.
Thus, $24=3+(8-1) d$, so that $d=3$.
Therefore, A1 = $a+d=3+3=6$;
$A_2=a+2 d=3+2 \times 3=9$
$A_3=a+3 d=3+3 \times 3=12$
$A_4=a+4 d=3+4 \times 3=15$
$A_5=a+5 d=3+5 \times 3=18$
$A_6=a+6 d=3+6 \times 3=21$
Therefore,six numbers between 3 and 24 are 6, 9, 12, 15, 18 and 21.
View full question & answer→Question 191 Mark
The income of a person is ₹3,00,000, in the first year and he receives an increase of ₹10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.
AnswerHere, we have an A.P. with a = 3,00,000, d = 10,000, and n = 20.
Using the sum formula, we obtain
$\mathrm{S}_{20}=\frac{20}{2}[600000+19 \times 10000]=10(790000)=79,00,000$
Therefore, the person received ₹79,00,000 as the total amount at the end of 20 years.
View full question & answer→Question 201 Mark
The sum of n terms of two Arithmetic Progression are in the ratio (3n + 8) : (7n + 15). Find the ratio of their $12^{th}$ terms.
AnswerLet $a_1, a_2$ and $d_1, d_2$ are the first term and common difference of two A.P.'S respectively.
According to question, $\frac{{\frac{n}{2}}}{{\frac{n}{2}}}\frac{{\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} = \frac{{3n + 8}}{{7n + 15}}$ .... (I)
$\frac { 12 \text { th term of Ist } \mathrm { A } . \mathrm { P } } { 12 \mathrm { th } \text { term of } 2 \mathrm { ndA.P } } = \frac { a _ { 1 } + 11 d _ { 1 } } { a _ { 2 } + 11 d _ { 2 } }$
put n = 23 in eq (i)
$\frac { 2 a _ { 1 } + 22 d _ { 1 } } { 2 a _ { 2 } + 22 d _ { 2 } } = \frac { 3 \times 23 + 8 } { 7 \times 23 + 15 }$
$\frac { a _ { 1 } + 11 d _ { 1 } } { a _ { 2 } + 11 d _ { 2 } } = \frac { 7 } { 16 }$
View full question & answer→Question 211 Mark
If the sum of n terms of an A.P. is $n \mathrm{P}+\frac{1}{2} n(n-1) \mathrm{Q}$, where P and Q are constants, find the common difference.
AnswerLet $a_1, a_2, \ldots a_n$ be the given A.P. Then
$\mathrm{S}_n=\mathrm{a}_1+\mathrm{a}_2+\mathrm{a}_3+\ldots+\mathrm{a}_{n-1}+\mathrm{a}_n=n \mathrm{P}+\frac{1}{2} n(n-1) \mathrm{Q}$
$\text { Thus, } \mathrm{S} 1=\mathrm{a}_1=\mathrm{P}, \mathrm{~S}_2=\mathrm{a}_1+\mathrm{a}_2=2 \mathrm{P}+\mathrm{Q}$
So that $\mathrm{a}_2=\mathrm{S}_2-\mathrm{S}_1=\mathrm{P}+\mathrm{Q}$
Therefore, the common difference is given by $d=a_2-a_1=(P+Q)-P=Q$.
View full question & answer→Question 221 Mark
In an A.P. if $m^{th}$ term is n and the $n^{th}$ term is m, where m $\ne$ n, find the $p^{th}$ term.
AnswerWe have $a_m=a+(m-1) d=n$,.....(i) and $a_n=a+(n-1) d=m$.....(ii)
Solving (i) and (ii), we obtain
$(m-n) d=n-m$, or $d=-1$, .....(iii)
and $\mathrm{a}=\mathrm{n}+\mathrm{m}-1$.....(iv)
Thus, $a_p=a+(p-1) d$ $=n+m-1+(p-1)(-1)=n+m-p$
Thus, the $\mathrm{p}^{\text {th }}$ term is $\mathrm{n}+\mathrm{m}-\mathrm{p}$.
View full question & answer→Question 231 Mark
Let the sequence $a_n$ be defined as follow: $a_1=1, a_n=a_{n-1}+2$ for $n \geq 2$
Find first five terms and write corresponding series.
AnswerGiven,
$a_1=1, a_2=a_1+2=1+2=3, a_3=a_2+2=3+2=5$
$a_4=a_3+2=5+2=7, a_5=a_4+2=7+2=9$
Thus, the first five terms of the sequence are $1,3,5,7$ and 9
The corresponding series is $1+3+5+7+9+\ldots$
View full question & answer→Question 241 Mark
If p, q, r are in G. P and the equation $px^2 + 2qx + r = 0$ and $dx^2 + 2ex + f = 0$ have a common root, that show that $\frac { d } { p } , \frac { e } { q } , \frac { f } { r }$are in A. P.
Answer$px^2 +2qx + r = 0$ has root
given by
$x = \frac { - 2 q \pm \sqrt { 4 q ^ { 2 } - 4 r p } } { 2 p }$
since p, q, r in G.P.
$q^2 = pr$
$x = \frac { - q } { p }$
but $\frac { - q } { p }$ is also root of
$dx^2 + 2ex + f = 0$
$d \left( \frac { - q } { p } \right) ^ { 2 } + 2 e \left( \frac { - q } { p } \right) + f = 0$
$dq^2 - 2eqp + fp^2 = 0$
$\div$ by $pq^2$ and using $q^2 = pr$
$\frac { d } { p } - \frac { 2 e } { q } + \frac { f p } { p r } = 0$
$\frac { 2 e } { q } = \frac { d } { p } + \frac { f } { r }$
Hence $\frac { d } { p } , \frac { e } { q } , \frac { f } { r }$ are in A.P
View full question & answer→Question 251 Mark
If a, b, c, d and p are different real numbers such that $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le$ 0, then show that a, b, c and d are in G.P.
AnswerHere, it is given that
$\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p+\left(b^2+c^2+d^2\right) \leq 0$...(i)
But L.H.S.
$=\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2-2 b c p+c^2\right)+\left(c^2 p^2-2 c d p+d^2\right),$
which gives $(a p-b)^2+(b p-c)^2+(c p-d)^2 \geq 0 \ldots$ (ii)
Since the sum of squares of real numbers is non negative, thus, from (i) and (ii), we have,
$(a p-b)^2+(b p-c)^2+(c p-d)^2=0$
or $a p-b=0, b p-c=0, c p-d=0$
$\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=\mathrm{p}$
Therefore, $a, b, c$ and $d$ are in G.P.
View full question & answer→Question 261 Mark
If a, b, c are in G. P and $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$ prove that x, y, z are in A. P.
Answer$\text { Let } a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{2}}=k$
$a=k^{\mathrm{x}}, \mathrm{~b}=\mathrm{k}^{\mathrm{y}}, \mathrm{c}=\mathrm{k}^z$
since $a, b, c$ are in G. P
$\mathrm{b}^2=\mathrm{ac}$
$\mathrm{k}^{2 \mathrm{y}}=\mathrm{k}^{\mathrm{x}+\mathrm{z}}$
$2 \mathrm{y}=\mathrm{x}+\mathrm{z}$
$\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in $\mathrm{A} . \mathrm{P}$
View full question & answer→Question 271 Mark
If $p^{\text {th }}, q^{\text {th }}, r^{\text {th }}$ and $s^{\text {th }}$ terms of an A.P. are in G.P, then show that $(p-q),(q-r),(r-s)$ are also in G.P.
Answer$\text { Given, } a_p=a+(p-1) d \ldots \text { (i) }$
$a_q=a+(q-1) d \ldots \text { (ii) }$
$a_r=a+(r-1) d \ldots \text { (iii) }$
$a_s=a+(s-1) d \ldots \text { (iv) }$
Given that $a_p, a_q, a_r$ and $a_s$ are in G.P, therefore
$\frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_q-a_r}{a_p-a_q}=\frac{q-r}{p-q} \ldots(\mathrm{v})$
Similarly $\frac{a_r}{a_q}=\frac{a_s}{a_r}=\frac{a_r-a_s}{a_q-a_r}=\frac{r-g}{q-r} \ldots$ (vi)
Therefore, by (v) and (vi)
$\frac{q-r}{p-q}=\frac{r-s}{q-r} \text {, i.e. } \mathrm{p}-\mathrm{q}, \mathrm{q}-\mathrm{r} \text { and } \mathrm{r}-\mathrm{s} \text { are in G.P }$
View full question & answer→Question 281 Mark
Find the sum to n terms of the series whose n th term is n (n+3).
AnswerGiven that $a_n = n (n + 3) = n^2 + 3n$
Therefore,the sum to n terms is given by
$\mathrm{S}_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k^{2}+3 \sum_{k=1}^{n} k$
= $\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}$ = $\frac{n(n+1)(n+5)}{3}$
View full question & answer→Question 291 Mark
What is the 20th term of the sequence defined by $a_n=(n-1)(2-n)(3+n) ?$
AnswerHere, $\mathrm{n}=20$,
$\therefore a_{20}=(20-1)(2-20)(3+20)$
$=19 \times(-18) \times(23)=-7866$
View full question & answer→Question 301 Mark
Find the sum to n terms of the series 5 + 11 + 19 + 29 + 41 + ........
Answer$S_n=5+11+19+29+\cdots+a_{n-1}+a_n$
$S_n=5+11+19+\cdots+a_{n-1}+a_{n-2}+a_n$
On subtracting
$0=5+[6+8+10+12+\cdots+(\mathrm{n}-1) \text { terms }]-\mathrm{a}_n$
$a_n=5+\frac{(n-1)[12+(n-2) \times 2]}{2}$
$=5+(\mathrm{n}-1)(\mathrm{n}+4)$
$=\mathrm{n}^2+3 \mathrm{n}+1$
${S_n} = \sum\limits_{k = 1}^n {{k^2}} + 3\sum\limits_{k = 1}^n k + n$
$= \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { 3 n ( n + 1 ) } { 2 } + n$
$= \frac { n ( n + 2 ) ( n + 4 ) } { 3 }$
View full question & answer→Question 311 Mark
If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.
Answer$\text { Given that A.M. }=\frac{a+b}{2}=10 \ldots . . \text { (i) }$
$\text { and G.M. }=\sqrt{a b}=8 \ldots . . . \text { (ii) }$
From (i) and (ii), we obtain
$a+b=20 \ldots . . . . \text { (iii) }$
$a b=64 \ldots . .(\text { (iv) }$
Substituting the value of $a$ and $b$ from (3), (4) in the identity $(a-b)^2=(a+b)^2-4 a b$, we obtain,
$(a-b)^2=400-256=144$
or $a-b= \pm 12$.....(v)
Solving (iii) and (v), we
$a=4, b=16 \text { or } a=16, b=4$
Thus, the numbers a and b are 4,16 or 16,4 respectively
View full question & answer→Question 321 Mark
Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.
AnswerSuppose $G_1, G_2, G_3$ be three numbers between 1 and 256 such that $1, G_1, G_2, G_3, 256$ is a G.P.
Thus, $256=r^4$ giving $r= \pm 4$ (Taking real roots only)
For $r=4$, we have $G_1=a r=4, G_2=a r^2=16, G_3=a r^3=64$
Similarly, for $r=-4$, numbers are $-4,16$ and -64
Therefore, we can insert $4,16,64$ between 1 and 256 so that the resulting sequences are in G.P.
View full question & answer→Question 331 Mark
A person has 2 parents, 4 grandparents, 8 great grandparents and so on. Find the numbers of his ancestors during the ten generations preceding his own.
AnswerThe number of ancestors is: 2, 4, 8,16, ....
It is in GP since;
$\frac 42$ = 2
$\frac 84$ = 2
$\frac {16}8$ = 2 ...
Therefore, common ratio, r = 2
a = 2
and n = 10
$\because S_n = as r > 1$
$\therefore S_{10} = \frac { 2 \left( 2 ^ { 10 } - 1 \right) } { 2 - 1 }$
= $2 (2^{10} - 1) = 2 (1024 - 1)$
= 2 $\times$ 1023 = 2046
Hence, the numbers of ancestors preceding the person is 2046
View full question & answer→Question 341 Mark
Find the sum of the sequence 7, 77, 777, ........ to n.
Answer$\mathrm{S}_n=7+77+777+\ldots-+\mathrm{n} \text { terms }$
$=\frac{7}{9}[9+99+999+\cdots-\cdots+\boldsymbol{n} \text { terms }]$
$=\frac{7}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\cdots+n \text { terms }\right]$
$=\frac{7}{9}\left[\left(10^1+10^2+\cdots-+n \text { terms }\right)-(1+1+1 \ldots++n \text { terms })\right]$
$=\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]$
$=\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{9}-n\right]$
View full question & answer→Question 351 Mark
The sum of first three terms of a GP is $\frac { 13 } { 12 }$ and their product is - 1. Find the terms.
AnswerLet the three numbers be $\frac { a } { r }$, a and ar.
Their sum = $\frac { a } { r }$ + a +ar = $\frac { 13 } { 12 }$
$\Rightarrow$ $a \left[ \frac { 1 } { r } + 1 + r \right] = \frac { 13 } { 12 }$
$\Rightarrow$ $a \left[ \frac { 1 + r + r ^ { 2 } } { r } \right] = \frac { 13 } { 12 }$
$\Rightarrow a\left(1+r+r^2\right)=\frac{13}{12} r \ldots(\mathrm{i})$
Their product $=\frac{a}{r} \times a \times a r=-1 \Rightarrow a^3=-1$
$\Rightarrow \mathrm{a}=-1$ [taking cube root on both sides] ...(ii)
On putting the value of a in Eq. (i), we get
$(-1)\left[1+r+r^2\right]=\frac{13}{12} r$
$\Rightarrow-1-r-r^2=\frac{13}{12} r$
$\Rightarrow-12-12 r-12 r^2=13 r$
$\Rightarrow 12 r^2+25 r+12=0$
$\Rightarrow 12 r^2+16 r+9 r+12=0$
$\Rightarrow 4 r(3 r+4)+3(3 r+4)=0$
$\Rightarrow(4 r+3)(3 r+4)=0$
$\Rightarrow \text { Either } 3 r+4=0 \text { or } 4 r+3=0$
$\Rightarrow$ r = $-\frac { 4 } { 3 }$ or r = $\frac { - 3 } { 4 }$
When a = - 1 and r = - $\frac { 4 } { 3 }$, then the numbers are
$\frac { - 1 } { - 4 / 3 } , - 1 , - 1 \times \frac { - 4 } { 3 } i . e . , \frac { 3 } { 4 } , - 1 , \frac { 4 } { 3 }$
And when a = - 1 and r = - $\frac { 3 } { 4 }$, then the numbers are
$\frac { - 1 } { - 3 / 4 } , - 1 , - 1 \times \frac { - 3 } { 4 } \text { i.e., } \frac { 4 } { 3 } , - 1 , \frac { 3 } { 4 }$
View full question & answer→Question 361 Mark
How many terms of GP 3, $\frac { 3 } { 2 } , \frac { 3 } { 4 }$, ... are needed to give the sum $\frac { 3069 } { 512 }$?
AnswerGiven GP is 3, $\frac { 3 } { 2 } , \frac { 3 } { 4 }$, ...
Here, a = 3, r = $\frac { 3 } { 2 } \div$ 3 = $\frac { 1 } { 2 }$
Let n be the number of terms needed.
Then, $S_n$ = $\frac { 3069 } { 512 }$
$\Rightarrow$ $\frac { a \left( 1 - r ^ { n } \right) } { 1 - r } = \frac { 3069 } { 512 }$ [$\because$r < 1]
$\Rightarrow$ $\frac { 3 \left\{ 1 - \frac { 1 } { 2 ^ { n } } \right\} } { 1 - \frac { 1 } { 2 } } = \frac { 3069 } { 512 }$
$\Rightarrow$ $6 \left( 1 - \frac { 1 } { 2 ^ { n } } \right) = \frac { 3069 } { 512 }$
$\Rightarrow$ 1 - $\frac { 1 } { 2 ^ { n } } = \frac { 3069 } { 3072 }$
$\Rightarrow$ $\frac { 1 } { 2 ^ { n } }$ = 1 - $\frac { 3069 } { 3072 } = \frac { 3072 - 3069 } { 3072 }$
$\Rightarrow$ $\frac { 1 } { 2 ^ { n } } = \frac { 3 } { 3072 } = \frac { 1 } { 1024 }$
$\Rightarrow 2^n$ = 1024 $\Rightarrow 2^n = 2^{10}$
On comparing the powers, we get
n = 10
Hence, 10 terms are needed to give the sum $\frac { 3069 } { 512 }$.
View full question & answer→Question 371 Mark
Find the sum of first n terms and the sum of first 5 terms of the geometric series $1 + \frac { 2 } { 3 } + \frac { 4 } { 9 } + - --$
Answera = 1, r = $\frac{2}{3}$
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$=\frac{1\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{2}{3}}$
$=3\left[1-\left(\frac{2}{3}\right)^{n}\right]$
$S_{5}=3\left[1-\left(\frac{2}{3}\right)^{5}\right]=\frac{211}{81}$
View full question & answer→Question 381 Mark
Write the first three terms of $a_{n}=\frac{n-3}{4}$
AnswerHere, $a_{n}=\frac{n-3}{4}$. Thus, $a_{1}=\frac{1-3}{4}=-\frac{1}{2}, a_{2}=-\frac{1}{4}, a_{3}=0$
Therefore the first three terms are $-\frac{1}{2},-\frac{1}{4}$ and 0.
View full question & answer→Question 391 Mark
In a G.P., the $3^{\text {rd }}$ term is 24 and the $6^{\text {th }}$ term is 192 . Find the $10^{\text {th }}$ term.
AnswerGiven, $a_3=a r^2=24 \ldots$ (i)
and $a_6=a r^5=192 \ldots$...ii)
Dividing (ii) by (i), we get $r =2$.
Putting, $r =2$ in (i), we get $a =6$.
Hence $a_{10}=6(2)^9=3072$
View full question & answer→Question 401 Mark
Write the first three terms of $a_n = 2n + 5$
AnswerHere $a_n=2 n+5$
Substituting $n =1,2,3$, we obtain
$a_1=2(1)+5=7, a_2=9, a_3=11$
Thus, the required terms are 7,9 and 11.
View full question & answer→Question 411 Mark
Which term of the G.P., 2,8,32, ... up to n terms is 131072?
AnswerLet 131072 be the $n ^{\text {th }}$ term of the given G.P. Here $a =2$ and $r =4$.
Thus, $131072=a_n=2(4)^{n-1}$ or $65536=4^{n-1}$
This gives $4^8=4^{n-1}$.
So that $n -1=8$, i.e., $n =9$. Therefore, 131072 is the $9^{\text {th }}$ term of the G.P.
View full question & answer→