5 Marks Questions

Question 15 Marks

If the sum of first $6$ terms of an $A.P$. is $36$ and that of the first $16$ terms is $256,$ find the sum of first $10$ terms.

Answer

$($We have the sum of first $n$ terms of an $AP) ,$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Given,
$36 =\frac{6}{2}[2 a+(6-1) d]$
$12 =2 a+5 d \ldots \ldots(1)$
$256 =\frac{16}{2}[2 a+(16-1) d]$
$32 =2 a+15 d \ldots \ldots(2)$
Subtracting $, (1)$ from $(2)$
$32-12 =2 a+15 d-(2 a+5 d)$
$\Rightarrow 20 =10 d$
$\Rightarrow d$
$\Rightarrow =2$
Substituting for $d$ in $(1),$
$12 =2 a+5(2)=2(a+5)$
$\Rightarrow 6 =a+5$
$\Rightarrow a =1$
$\therefore$ The sum of first $10$ terms of an $AP ,$
$S_{10}=\frac{10}{2}[2(1)+(10-1) 2]$
$S_{10}=5[2+18]$
$S_{10}=100$
This is the sum of the first $10$ terms

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Question 25 Marks

The ratio of the $11^{\text {th }}$ term to the $18^{\text {th }}$ term of an $A.P$. is $2: 3$. Find the ratio of the $5^{\text {th }}$ term to the $21^{\text {st }}$ term. Also, find the ratio of the sum of first $5$ terms to the sum of first $21$ terms.

Answer

Let a and $d$ be the first term and common difference of given $AP$. it is given that
$\frac{a_{11}}{a_{18}}=\frac{2}{3}$
$\Rightarrow \frac{a+10 d}{a+17 d}=\frac{2}{3}$
$\Rightarrow 3(a+10 d)=2(a+17 d)$
$\Rightarrow 3 a+30 d=2 a+34 d$
$\Rightarrow a=4 d$
Ratio of $5^{\text {th }}$ term to the $21^{\text {tt }}$ term is
$\frac{a_5}{a_{21}}=\frac{a+4 d}{a+20 d} \ldots \ldots(1)$
putting $a =4 d$ in $(1),$ we get
$\frac{a_5}{a_{21}}=\frac{4 d+4 d}{4 d+20 d}$
$\frac{a_5}{a_{21}}=\frac{8 d}{24 d}$
$\frac{a_5}{a_{21}}=\frac{1}{3}$
Ratio of $S_5$ to the $S_{21}$ is
$\frac{S_5}{S_{21}}=\frac{\frac{5}{2}[2 a+4 d]}{\frac{21}{2}[2 a+20 d]}$
$\left(\because S_n=\frac{n}{2}(2 a+(n-1) d)\right)$
$\frac{S_5}{S_{21}}=\frac{5[2 a+4 d]}{21[2 a+20 d]} \ldots \ldots(2)$
Putting $a=4 d$ in $(ii),$ we get
$\frac{S_5}{S_{21}}=\frac{5[2(4 d)+4 d]}{21[2(4 d)+20 d]}$
$\frac{S_5}{S_{21}}=\frac{5[12 d]}{21[28 d]}$
$\frac{S_5}{S_{21}}=\frac{60 d}{588 d}$
$\frac{S_5}{S_{21}}=\frac{5}{49}$

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Question 35 Marks

The first term of an A.P is 5 , the last term is 45 and the sum is 400.
Find the number of terms and the common difference.

Answer

First term $a=5$, last term $1=45$ and $S_n=400$
Since,
$
S_{n}=\frac{n}{2}(a+1)
$
$
\begin{array}{ll}
\Rightarrow & 400=\frac{n}{2}(5+45) \\
\Rightarrow & 400=\frac{n}{2}(50) \\
\Rightarrow & 400=n(25)
\end{array}
$
$\begin{array}{lrl}\Rightarrow & n & =16 \\ \text { Since, } & 1 & = a +( n -1) d \\ \Rightarrow & 45 & =5+(16-1) d \\ \Rightarrow & 45-5 & =(15) d \\ \Rightarrow & 40 & =15 d \\ \Rightarrow & d & =\frac{40}{15} \\ \Rightarrow & d & =\frac{8}{3}\end{array}$

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Question 45 Marks

Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 , respectively

Answer

Here, $\quad a_2=14$ and $a_s=18$
Common difference,(1) Now
$
\begin{aligned}
d & =a_{s}-a_2 \\
& =18-14=4 \\
a_2 & =a+d=14 \\
a+4 & =14 \\
a & =10
\end{aligned}
$
Now, sum of 51 terms
$
\begin{aligned}
S_{n} & =\frac{n}{2}(2 a+(n=1) d) \\
S_{51} & \left.=\frac{51}{2}(210)+(51-1) 4\right) \\
& =\frac{51}{2}(20+200) \\
& =\frac{51 \times 220}{2} \\
& =51 \times 110 \\
& =5610 \\
S_{51} & =5610
\end{aligned}
$

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Question 55 Marks

The ratio of the sums of first $m$ and first $n$ terms of an A. P. is $m^2: n^2$.
Show that the ratio of its $m^{th}$ and $n^{th}$ terms is $(2 m-1):(2 n-1)$.

Answer

We know that sum of first $n$ terms of an A.P is given by the formula
$S_n=\frac{n}{2}\{2 a+(n-1) d\}$
Similarly sum of first $m$ terms of the same A.P. will be
$S_m=\frac{m}{2}\{2 a+(m-1) d\}$
According to the Question
$\begin{array}{l}\frac{S_m}{S_n}=\frac{m^2}{n^2} \\\Rightarrow \frac{S_m}{S_n}=\frac{\frac{m}{2}(2 a+(m-1) d)}{\frac{n}{2}(2 a+(n-1) d)}=\frac{m^2}{n^2} \\\Rightarrow \frac{m\{2 a+(m-1) d\}}{n\{2 a+(n-1) d\}}=\frac{m^2}{n^2} \\\Rightarrow \frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n} \\\Rightarrow n \times(2 a+(m-1) d)=m \times(2 a+(n-1) d) \\\Rightarrow 2 n a+n(m-1) d=2 m a+m(n-1) d\end{array}$
$\begin{array}{l}\Rightarrow 2 a n+m n d-n d=2 a m+m n d-m d \\\Rightarrow 2 a n-n d=2 a m-m d \\\Rightarrow m d-n d=2 a m-2 a n \\\Rightarrow(m-n) d=2 a(m-n) \\\Rightarrow d=2 a\end{array}$
We know that $n^{\text {th }}$ term of an A.P is given by
$a_n=a+(n-1) d\ldots\ldots \text {(i)}$
Similarly term of the A.P will be given by
$a_m=a+(m-1) d\ldots\ldots \text {(ii)}$
Ratio of $m^{\text {th }}$ to the $n^{\text {th }}$ term will be,
$\frac{a_m}{a_n}=\frac{a+(m-1) d}{a+(n-1) d}$
Put $d=2 \alpha$
$\begin{array}{l}\Rightarrow \frac{a_m}{a_n}=\frac{a+(m-1) 2 a}{a+(n-1) 2 a} \\\Rightarrow \frac{a_m}{a_n}=\frac{a+2 a m-2 a}{a+2 a n-2 a} \\\Rightarrow \frac{a_m}{a_n}=\frac{a+2 a m-2 a}{a+2 a n-2 a} \\\Rightarrow \frac{a_m}{a_n}=\frac{2 a m-a}{2 a n-a}=\frac{a(2 m-1)}{a(2 n-1)} \\\Rightarrow \frac{a_m}{a_n}=\frac{2 m-1}{2 n-1}\end{array}$
So, the ratio of its $m^{\text {th }}$ and $n^{\text {th }}$ terms is $(2 m-1)$ : $(2 n-1)$
Hence proved.

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Question 65 Marks

If the ratio of the sum of the first $n$ terms of two A.Ps is $(7 n+1):(4 n+27)$, then find the ratio of their $9^{\text {th }}$terms.

Answer

Let $a_1, a_2$ be first terms of the two A.Ps.
Let $d_1, d_2$ be common differences of the two A.Ps. Sum of the n terms are
$\begin{array}{l}S_n=\frac{n}{2}\left\{2 a_1+(n-1) d_1\right\} \text { and } S_n^{\prime}=\frac{n}{2}\left\{2 a_2+(n-1) d_2\right\} \\\therefore \frac{S_n}{S_n}=\frac{\frac{n}{2}\left\{2 a_1+(n-1) d_1\right\}}{\frac{n}{2}\left\{2 a_2+(n-1) d_2\right\}}=\frac{2 a_1(n-1) d_1}{2 a_2+(n-1) d_2}\end{array}$
Given,
$\frac{S_n}{S_n{ }^{\prime}}=\frac{7 n+1}{4 n+27}$
$\Rightarrow \frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2}=\frac{7 n+1}{4 n+27}$
Divide both numerator and denominator on the LHS by 2 .
As, $a_{19}=a+8 d$
Comparing with $\left\{a+\frac{(n-1)}{2} d\right\}$ we get,
$\begin{array}{l}a+\frac{(n-1)}{2} d=a+8 d \\\Rightarrow \frac{(n-1)}{2}=8\end{array}$
After solving we get,
$\Rightarrow n=17$
Therefore,
$\frac{a_9}{a_9{ }^{\prime}}=\frac{\left\{a_1+8 d_1\right\}}{\left\{a_2+8 d_2\right\}}=\frac{7(17)+1}{4(17)+27}=\frac{120}{95}=\frac{24}{19}$
Therefore the ratio of the $9^{\text {th }}$ terms of the two A.Ps is $\frac{24}{19}$.

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Question 75 Marks

Show that $a_1, a_2, a_3, \ldots \ldots . ., a_n$ form A.P. where $a_n$ is defined as $a _{ n }=9-5 n$.

Answer

Given $a_n=9-5 n$
$\begin{array}{l}a_1=9-5(1)=4 \\a_2=9-5(2)=-1 \\a_3=9-5(3)=-6\end{array}$
Difference between, $a_2$ and $a_1$
$\begin{array}{l}=a_2-a_1 \\=-1-4=-5\end{array}$
Difference between $a_3$ and $a_2$
Also, $a_n=9-5 n$ can be written as
$\begin{array}{l}a_n=4+5-5 n \\\Rightarrow a_n=4+(1-n)(5)\end{array}$
$\Rightarrow a_n=4+(n-1)(-5)$ which is the standard form of the $n^{\text {th }}$ term of an A.P.
Also, as the difference is same between these terms, hence, they form an A.P.
Hence proved.

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Question 85 Marks

Ramkali required Rs 2500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.What value is generated in the above situation?

Answer

Initial saving for first week is Rs 100 and increases every week by Rs 20, this forms an A.P., where the first term is $a=100$ and difference, $d=20$, Now to find if required total saving is Rs 2500 after 12 weeks, we need to find the sum of this AP for 12 terms. So, $n=12$
$\begin{array}{l}\text { Using } S_n=\frac{n}{2}(2 a+(n-1) d) \\S_{12}=\frac{12(2 \times 100+(12-1) 20)}{2} \\S_{12}=6(200+(11) 20) \\S_{12}=6(200+220) \\S_{12}=6(420)=2520\end{array}$
Hence we can say that after 12 weeks of saving Ramkali will generate Rs 2520, which is more than 2500.
So she will be able to send her daughter to school after 12 weeks.

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Question 95 Marks

Find the $60^{\text {th }}$ term of the $\text {AP}$ $ 8,10,12, \ldots$, if it has a total of terms and hence find the sum of its last 10 terms.

Answer

The given $\text {AP}$ is $8,10,12, \ldots,$
First term, $a=8$
Common difference, $d=10-8=2$
$\begin{array}{l}a_n=a+(n-1) d \\a_{60}=8+(60-1) 2 \\a_{60}=8+59 \times 2 \\a_{60}=126 \\S_n=\frac{n}{2}\{2 a+(n-1) d\}\end{array}$
Sum of first 60 terms
$\begin{array}{l}S_{60}=\frac{60}{2}\{2 \times 8+(60-1) 2\} \\S_{60}=30(16+59 \times 2) \\S_{60}=4020\end{array}$
Sum of first 50 terms
$\begin{array}{l}S_{50}=\frac{50}{2}\{2 \times 8+(50-1) 2\} \\S_{50}=25(16+49 \times 2) \\S_{50}=2850\end{array}$
Sum of last 10 terms is $S_{60}-S_{50}$
$=4020-2850=1170$
Hence, the sum of the last 10 terms is 1170 .

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Question 105 Marks

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. Which value is shown in this Question?

Answer

It is given that number of trees planted by each class is double the class in which they are studying and each class has two sections.
Number of trees planted by class $1=2 \times 2=4$
Number of trees planted by class $2=2 \times 4=8$
Number of trees planted by class $3=2 \times 6=12$ and so on.
Thus it forms an A.P. as $4,8,12, \ldots .48$
$\begin{array}{l}\text { Number of terms = total classes in school } \\\quad=n=12\end{array}$
Sum of n terms of an A.P is given by;
$S_n=\frac{n}{2}(a+l)$
So, sum of 12 terms of an A.P. $=S_{12}=\frac{12}{2}(4+48)$
$\begin{array}{l}\Rightarrow S_{12}=6(4+48)=24+288 \\\Rightarrow S_{12}=312\end{array}$
Hence, 312 trees were planted by the students.

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Question 115 Marks

In an $A P$ of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565 . Find the A.P.

Answer

Let ' $a$ ' and ' $d$ ' be the first term and the common difference of an A.P. respectively.
The $n^{\text {th }}$ term of an A.P is given by,
$a_n=a+(n-1) d$
And the sum of $n$ terms of an A.P,
$S_n=\frac{n}{2}[2 a+(n-1) d]$
So, sum of first 10 terms of an A.P is,
$\begin{array}{l}S_{10}=\frac{10}{2}[2 a+(10-1) d] \\\Rightarrow S_{10}=\frac{10}{2}[2 a+9 d] \\\Rightarrow 210=\frac{10}{2}[2 a+9 d] \\\Rightarrow 210=5[2 a+9 d] \\\Rightarrow 42=[2 a+9 d]\quad \quad \ldots \ldots(1)\end{array}$
Now, the $15^{\text {th }}$ term from the last $=$ $(50-15+1)^{t h}=36^{t h}$ term from the beginning
$a_{36}=a+35 d$
Thus, sum of the last 15 terms
$\begin{array}{l}=\frac{15}{2}\left[2 a_{36}+(15-1) d\right] \\\Rightarrow \frac{15}{2}[2(a+35 d)+(15-1) d] \\\Rightarrow \frac{15}{2}[2(a+35 d)+14 d]\end{array}$
$\begin{array}{l}\Rightarrow \frac{15}{2}[2(a+35 d+7 d)] \\\Rightarrow 15[a+35 d+7 d] \\\Rightarrow 15[a+42 d] \\
\Rightarrow 2565=15[a+42 d] \\\Rightarrow 171=[a+42 d]\quad \quad \ldots \ldots(2)\end{array}$
Solving (1) and (2), we get,
$a=3, d=4$
Hence, the required A.P is, $3,7,11,15$,$\ldots\ldots$ $199$

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Question 125 Marks

Find the common difference of an A.P. whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

Answer

Let the common difference of the A.P be $d$. First term is 5 .
Therefore, the A.P can be written as
$5,5+d, 5+2 d, 5+3 d, 5+4 d, 5+5 d, 5+6 d, 5+7 d$
The above sequence represents the first 8 terms of the A.P.
Now, the sum of the first 4 terms of the A.P
$=5+5+d+5+2 d+5+3 d=20+6 d$
And, the sum of the next four terms of the A.P
$=5+4 d+5+5 d+5+6 d+5+7 d=20+22 d$
Now, it is given that
Sum of the first 4 terms
$\begin{aligned}& =\frac{1}{2} \times \text { sum of the next four terms } \\\Rightarrow & 20+6 d=\frac{1}{2} \times(20+22 d) \\\Rightarrow & 20+6 d=10+11 d \\\Rightarrow & 11 d-6 d=20-10 \\\Rightarrow & 5 d=10 \\\Rightarrow & d=2\end{aligned}$
Therefore, the common difference of the A.P is 2 .

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Question 135 Marks

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is $7: 15$. Find the numbers.

Answer

Assume that the four consecutive numbers are $(a-3 d),(a-d),(a+d)$ and $(a+3 d)$
$\begin{array}{l}(a-3 d)+(a-d)+(a+d)+(a+3 d)=32 \\\Rightarrow 4 a=32 \\\Rightarrow a=8\end{array}$
Now the numbers are $(8-3 d),(8-d),(8+d)$ and $(8+3 d)$.
According to the Question
$\begin{array}{l}\frac{(8-3 d)(8+3 d)}{(8-d)(8+d)}=\frac{7}{15} \\\Rightarrow \frac{\left(8^2-9 d^2\right)}{\left(8^2-d^2\right)}=\frac{7}{15}\end{array}$
$\begin{array}{l}\Rightarrow \frac{\left(64-9 d^2\right)}{\left(64-d^2\right)}=\frac{7}{15} \\\Rightarrow 960-135 d^2=448-7 d^2 \\\Rightarrow 128 d^2=512\end{array}$
Divide both sides by 128 .
$\begin{array}{l}\Rightarrow d^2=4 \\\Rightarrow d= \pm 2\end{array}$
As the terms are consecutive hence the difference cannot be negative
Hence $d=2$
And the four consecutive terms are:
$(8-3 \times 2),(8-2),(8+2)$ and $(8+3 \times 2)$.
Or $2,6,10,14$.

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Question 145 Marks

If the sum of the first $p$ terms of an A.P. is the same as the sum of its first q terms (where $p \neq q$ ), then show that the sum of first $(p+q)$ terms is zero

Answer

Let $a$ and $d$ be the first term and common difference respectively of the given A.P.
using, $Sn =\frac{ n }{2}[2 a +( n -1) d ]$
$Sp = Sq$ （given）
$\begin{array}{l}\frac{p}{2}[2 a+(p-1) d]=\frac{q}{2}[2 a+(q-1) d] \\\Rightarrow 2 ap+p(p-1) d=2 aq+q(q-1) d \\\Rightarrow 2 a(p-q)+\{p(p-1)-q(q-1)\} d=0 \\\Rightarrow 2 a(p-q)+\left\{p^2-p-q^2+q\right\} d=0 \\\Rightarrow 2 a(p-q)+\left\{\left(p^2-q^2\right)-(p-q)\right\} d=0 \\\Rightarrow 2 a(p-q)+\{(p+q)(p-q)-(p-q\} d=0 \\\Rightarrow(p-q)[2 a+\{(p+q)-1\} d\}=0 \\\Rightarrow 2 a+(p+q-1) d=0 \quad \quad[\because p \neq q]\quad \quad \ldots \ldots\text {(i)}\end{array}$

$\begin{aligned}\text {Now}\quad S p+q & =\frac{p+q}{2}[2 a+(p+q-1) d] \\& =\frac{p+q}{2} \times 0 \quad \text { [from equation(i)] } \\& =0\end{aligned}$
Thus, $\quad S p+q=0$
Hence proved.

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