[5 marks sum]

Question 15 Marks

Find five numbers in A.P whose sum is $12 \frac{1}{2}$ and the ratio of the first to the last terms $2:3.$

Answer

Let the five numbers in A.P. be
$(a-2 d),(a-d), a,(a+d)$ and $(a+2 d)$.
Then, $(a-2 d)+(a-d)+a+(a+d)+(a+2 d)=12 \frac{1}{2}$
$\Rightarrow 5 a=\frac{25}{2} $
$ \Rightarrow a=\frac{5}{2}$
It is given that
$\frac{a-2 d}{a+2 d}=\frac{2}{3}$
$\Rightarrow 3 a-6 d=2 a+4 d $
$ \Rightarrow a=10 d $
$ \Rightarrow \frac{5}{2}=10 d $
$ \Rightarrow d=\frac{1}{4} $
$ \Rightarrow a=\frac{5}{2} \text { and } d=\frac{1}{4}$
Thus, We have
$a-2 d=\frac{5}{2}-2 \times \frac{1}{4}=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2$
$ a-d=\frac{5}{2}-\frac{1}{4}=\frac{10-1}{4}=\frac{9}{4} $
$ a=\frac{5}{2}$
$a+d=\frac{5}{2}+\frac{1}{4}=\frac{10+1}{4}=\frac{11}{4} $
$ a+3 d=\frac{5}{2}+2 \times \frac{1}{4}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$
Thus, the five numbers in $A . P=2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}$ and $3$
$=2,2.25,2.5,2.75$ and $3$

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Question 25 Marks

Divide $96$ into four parts which are in A.P and the ratio between product of their means to product of their extremes is $15:7.$

Answer

Let the four parts be $(a-3 d),(a-d),(a+d)$ and $(a+3 d)$
Then, $(a-3 d)+(a-d)+(a+d)+(a+3 d)=96$
$\Rightarrow 4 a=96$
$\Rightarrow a=24$
It is given that
$\frac{(a-d)(a+d)}{(a-3 d)(a+3 d)}=\frac{15}{7}$
$\Rightarrow \frac{a^2-d^2}{a^2-9 d^2}=\frac{15}{7}$
$\Rightarrow \frac{576-d^2}{576-9 d^2}=\frac{15}{7}$
$\Rightarrow 4032-7 d^2=8640-135 d^2$
$\Rightarrow 128 d^2=4608$
$\Rightarrow d^2=36$
$\Rightarrow d= \pm 6$
When $a=24, d=-6$
$a-3 d=24-3(6)=6 $
$ a-d=24-(-6)=18$
$ a+d=24+6=30$
$a+3 d=24+3(6)=42$
When $a=24, d=-6$
$a-3 d=24-3(-6)=42 $
$a-d=24-(-6)=30$
$a+d=24+(-6)=18$
$ a+3 d=24+3(-6)=6$
Thus , the four parts are $(6,18,30,42)$ or $(42,30,18,6)$

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Question 35 Marks

The sum of three consecutive terms of an A.P. is $21$ and the sum of their squares is $165.$ find these terms.

Answer

Let the treee consecutive terms in A.P.be a-d, a and $a+d.$
$\therefore(a-d)+a+(a+d)=21$
$\Rightarrow a=7........(1)$
Also,$(a-d)^2+a^2+(a+d)^2=165$
$\Rightarrow a^2+d^2-2 a d+a^2+a^2+d^2+2 a d=165 $
$ \Rightarrow 3 a^2+2 d^2=165$
$\Rightarrow 3 \times(7)^2+2 d^2=165......[From(1)]$
$ \Rightarrow 3 \times 49+2 d^2=165 $
$ \Rightarrow 147+2 d^2=165 $
$\Rightarrow 2 d^2=18 $
$\Rightarrow d^2=9 $
$ \Rightarrow d= \pm 3$
when $a=7$ and $d=3$
Required terms $=a-d, a$ and $a+d$
$=7-3,7,7+3 $
$ =4,7,10$
When $a=7$ and $d=-3$
$\text { Required terms }=a-d, a \text { and } a+d $
$ =7-(-3), 7,7+(-3) $
$=10,7,4$

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Question 45 Marks

If the sum of first $7$ terms of an A.P. is $49$ and that of its first $17$ terms is $289,$ find the sum of first n terms of the A.P.

Answer

Let the first term and the common difference of the given AP be a and d, respectively.
Sum of the first $7$ terms, $S_7 = 49$
We know
$S=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \frac{7}{2}(2 a+6 d)=49$
$\Rightarrow \frac{7}{2} \times 2(a+3 d)=49$
$\Rightarrow a+3 d=7 ....(1)$
Sum of the first $17$ terms, $S_{17} = 289$
$\Rightarrow \frac{17}{2}(2 a +16 d )=289$
$\Rightarrow \frac{17}{2} \times 2( a +8 d )=289$
$\Rightarrow a+8 d=\frac{289}{17}$
$\Rightarrow a+8 d=17....(2)$
Subtracting $(2)$ from $(1),$ we get
$5d = 10$
$5 d=10$
$\Rightarrow d=2$
Substituting the value of $d$ in $(1),$ we get
$a = 1$
Now,
Sum of the first n terms is given by
$S_n=\frac{n}{2}[2 a+(n-1) d] $
$ =\frac{n}{2}[2 \times 1+(n-1)]$
$ =n(1+n-1) $
$ =n^2$
Therefore, the sum of the first n terms of the AP is $n^2.$

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Question 55 Marks

The fourth term of an $A.P.$ is $11$ and the term exceeds twice the fourth term by $5$ the $A.P$ and the sum of first $50$ terms.

Answer

For an A.P $t_4=11$
$\Rightarrow a + 3d = 11 ....(1)$
Also , $t_8 - 2t_4 = 5$
$\Rightarrow (a + 7d) - 2 \times 11 = 5$
$\Rightarrow a + 7d - 22 = 5$
$\Rightarrow a + 7d = 27 ....(2)$
Substracting equation 1 from equation 2, we get
$a+3 d=11$
$a+7 d=27$
$\frac{----}{-4 d=-16}$
$d=\frac{-16}{-4}$
$\Rightarrow d=4$
substituting d = 4 in equation 1, we get
$a + 3d = 11$
$a + 3 \times 4 = 11$
$\Rightarrow a + 12 = 11$
$\Rightarrow a = -1$
$\therefore $ required $A.P = a, a + d, a + 2d, a + 3d,....$
$\therefore a = - 1$
$\therefore a + d = - 1 + 4 = 3$
$\therefore a + 2d = - 1 + 2(4) = - 1 + 8 = 7$
$\therefore a + 3d = - 1 + 3(4) = - 1 + 12 = 11$
Sum of first 50 terms of this A.p $=\frac{50}{2}(2 \times(-1)+49 \times 4)$
$= 25(- 2 + 196)$
$= 25 \times 194$
$= 4850$

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Question 65 Marks

$4$th term of an A.P is equal to $3$ times its first term and $7$th term exceeds twice the $3$rd time by I. Find the first term and the common difference.

Answer

The general term of an AP is given by $t_n=a+(n-1) d$
Now $t_4=3 \times a$
$\Rightarrow a + 3d = 3a$
$\Rightarrow 2a - 3d = 0 ...(1)$
$\text { Next } t_7-2 \times t_3=1 $
$ \Rightarrow a+6 d-2(a+2 d)=1 $
$ \Rightarrow a+6 d-2 a-4 d=1 $
$ \Rightarrow -a+2 d=1 \ldots . \text { (ii) }$
multiplying $(ii)$ by $2$ we get
$-2a + 4d = 2 ....(iii)$
Adding equation $(i)$ and $(iii)$ we get
$d = 2$
Substituting the value of d in $(ii)$ we get
$ -a+2 \times 2=1 $
$ \Rightarrow-a+4=1$
$\Rightarrow a=3 $
Hence $a =3$ and $d =2$

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Question 75 Marks

An A.P consists of $57$ terms of which $7$th term is $13$ and the last term is $108.$ Find the $45$th term of this A.P.

Answer

Number of terms $= n = 57$
$t_7=13$
$\Rightarrow a + 6d = 13 ....(i)$
Last term $=t_{57}=108$
$\Rightarrow a + 56d = 108 ....(ii)$
Substracting (i) from (ii) we get
$50d = 95$
$\Rightarrow d=\frac{95}{50}$
$\Rightarrow d=\frac{19}{10}$
Substituting value of d in (i) we get
$a+6 \times \frac{19}{10}=13$
$\Rightarrow a+\frac{57}{5}=13$
$\Rightarrow a=13-\frac{57}{5}=\frac{65-57}{5}=\frac{8}{5}$
$\Rightarrow>\text { General term }=t_n=\frac{8}{5}+(n-1) \times \frac{19}{10}$
$\Rightarrow t_{45}=\frac{8}{5}+44 \times \frac{19}{10}=\frac{8}{5}+\frac{418}{5}=\frac{426}{5}=85.2$

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Question 85 Marks

The sum of the $4^{\text {th }}$ and the $8^{\text {th }}$ terms of an $A.P$. is $24$ and the sum of the $6^{\text {th }}$ and the $10^{\text {th }}$ terms of the same $A.P.$ is $34$ . Find the first three terms of the $A.P.$

Answer

Let ' $a$ ' be the first term and ' $d$ ' be the common difference of the given A.P.
$ t_4+t_8=24 \text { (given) }$
$\Rightarrow(a+3 d)+(a+7 d)=24$
$\Rightarrow 2 a+10 d=24$
$\Rightarrow a+5 d=12 \ldots \text { (i) } $
And,
$ t_6+t_{10}=34 \text { (given) }$
$\Rightarrow(a+5 d)+(a+9 d)=34$
$\Rightarrow 2 a+14 d=34$
$\Rightarrow a+7 d=17 \ldots \text { (ii) } $
Subtracting (i) from (ii), we get
$2 d=5$
$\Rightarrow d=\frac{5}{2}$
$\Rightarrow a+5 \times \frac{5}{2}=12$
$\Rightarrow a+\frac{25}{2}=12$
Thus we have,
$1^{\text {st }}$ term $=-\frac{1}{2}$
$2^{n d}=a+d=-\frac{1}{2}+\frac{5}{2}=2$
$3^{r d}$ term $=a+2 d=-\frac{1}{2}+2 \times \frac{5}{2}=-\frac{1}{2}+5=\frac{9}{2}$

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Question 95 Marks

If $t_n$ represents $n ^{\text {th }}$ term of an A.P $t_2+t_5-t_3=10$ and $t_2+t_9=17$ Find its first term and its common difference.

Answer

Let the first term of an A.P be a and the common difference be d.
The general term of an A.P is given by $t_n=a+(n-1) d$
Now $t_1+t_5-t_3=10$
$\Rightarrow(a+d)+(a+4 d)-(a+2 d)=10 $
$\Rightarrow a+d+a+4 d-a-2 d=10$
$\Rightarrow a + 3d = 10 ...(i)$
Also $t_2+t_9=17$
$\Rightarrow (a + d) + (a + 8d) = 17$
$\Rightarrow2a + 9d = 17 ....(ii)$
Multiplying equation $(i)$ by $2$ we get
$2a + 6d = 20 ....(iii)$
Subtracting $(ii)$ from $(iii)$ we get
$-3d= 3$
$\Rightarrow d = -1$
Substituting value of d in (i) we get
$a + 3(-1) = 10$
$\Rightarrow a - 3 = 10$
$\Rightarrow a = 13$
Hence $a = 13$ and $d = -1$

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Mathematics [5 marks sum]

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