5 Marks Questions

Question 15 Marks

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

Answer

Let the first term and the common difference of the AP be a and d respectively.
According to the question,
Third term + seventh term = 6
$ \Rightarrow $ [a + (3 - 1)d] + [a + (7 - 1)d] = 6 = a + (n - 1)d
$ \Rightarrow $ (a + 2d) + (a + 6d) = 6 $ \Rightarrow $ 2a + 8d = 6
$ \Rightarrow $ a + 4d = 3 ..... (1)
Dividing throughout by 2 &
(third term) (seventh term) = 8
$ \Rightarrow $ (a + 2d) (a + 6d) = 8
$ \Rightarrow $ (a + 4d - 2d) (a + 4d + 2d) = 8
$ \Rightarrow $ (3 - 2d) (3 + 2d) = 8
$ \Rightarrow $ $9 - 4d^2 = 8$
$ \Rightarrow 4{d^2} = 1 \Rightarrow {d^2} = \frac{1}{4} \Rightarrow d + \pm \frac{1}{2}$
Case I, when $d = \frac{1}{2}$
Then from (1), $a + 4\left( {\frac{1}{2}} \right) = 3$
$ \Rightarrow $ a + 2 = 3 $ \Rightarrow $ a = 3 - 2 $ \Rightarrow $ a = 1
$\therefore $ Sum of first sixteen terms of the $AP = S_{16}$
$ = \frac{{16}}{2}[2a + (16 - 1)d]$ $\because {S_n} = \frac{n}{2}[2a + (n - 1)d]$
= 8[2a + 15d]
$ = 8[2(1) + 15(\frac{1}{2})]$
$ = 8[12 + \frac{{15}}{2}]$
$ = 8[\frac{{19}}{2}]$
$ = 4 \times 19 = 76$
Case II. When $d = - \frac{1}{2}$
Then from (1),
$a + 4\left( { - \frac{1}{2}} \right) = 3$
$ \Rightarrow $ a - 2 = 3 $ \Rightarrow $ a = 3 + 2 $ \Rightarrow $ a = 5
$\therefore $ Sum of first sixteen terms of the $AP = S_{16}$
$ = \frac{{16}}{2}[2a + (16 - 1)d]$ $\because {S_n} = \frac{n}{2}[2a + (n - 1)d]$
$ = 8[2a + 15d] = 8\left[ {2(5) + 15\left( { - \frac{1}{2}} \right)} \right] = 8\left[ {10 - \frac{{15}}{2}} \right] = 8\left[ {\frac{5}{2}} \right] = 20$

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Question 25 Marks

In an $AP: a_n = 4, d = 2, S_n = -14$, find n and a.

Answer

Here, $a_n = 4$
$d = 2$
$S_n = -14$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow $ 4 = a + (n - 1)d
$ \Rightarrow $ 4 = a + 2n - 2
$ \Rightarrow $ 4 + 2 = a + 2n
$ \Rightarrow $ 6 = a + 2n
$ \Rightarrow $ a + 2n = 6 ...... (1)
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow - 14 = \frac{n}{2}\left[ {2a + (n - 1)2} \right]$
$ \Rightarrow $ -14 = n[a + (n - 1)]
$ \Rightarrow $ -14 = n (a + n - 1)
$ \Rightarrow $ -14 = n (6 - n - 1) .......From (1), (a + 2n = 6 $ \Rightarrow $ a + n = 6 - n)
$ \Rightarrow $ -14 = n(-n + 5)
$ \Rightarrow $ $-14 = -n^2 + 5n$
$ \Rightarrow $ $n^2 - 7n + 2n - 14 = 0$
$ \Rightarrow $ n(n - 7) + 2(n - 7) = 0
$ \Rightarrow $ (n - 7) (n + 2) = 0
$ \Rightarrow $ n - 7 = 0 or n + 2 = 0
$ \Rightarrow $ n = 7 or n = -2
$ \Rightarrow $ n = - 2 is in admissible as n, being the number of terms, is a natural number.
$\therefore $ n = 7
Putting n = 7 in equation (1), we get
a + 2(7) = 6
$ \Rightarrow $ a + 14 = 6
$ \Rightarrow $ a = 6 - 14
$ \Rightarrow $ a = -8

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Question 35 Marks

In an AP: $a_3=15, S_{10}=125$, find $d$ and $a_{10}$.

Answer

Here, $a_3=15$
$S_{10}=125$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow a_3 = a + (3 - 1)d$
$ \Rightarrow a_3 = a + 2d$
$ \Rightarrow 15 = a + 2d$
$ \Rightarrow a + 2d = 15 ...... (1)$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{10}} = \frac{{10}}{2}\left[ {2a + (10 - 1)d} \right]$
$ \Rightarrow S_{10} = 5(2a + 9d)$
$ \Rightarrow 125 = 5(2a + 9d)$
$ \Rightarrow 25 = 2a + 9d$
$ \Rightarrow 2a + 9d = 25 ....... (2)$
Solving equation (1) and equation (2), we get
$a = 17$
$d = -1$
Now $an = a + (n - 1)d$
$ \Rightarrow a_{10} = a + (10 - 1)d$
$ \Rightarrow a_{10} = a + 9d$
$ \Rightarrow a_{10} = 17 + 9(-1)$
$ \Rightarrow a_{10} = 17 - 9$
$ \Rightarrow a_{10} = 8$

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Question 45 Marks

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig.). In how many rows are the 200 logs placed and how many logs are in the top row?

Answer

Self Learning

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Question 55 Marks

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown in Figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? $(Take\; \pi = \frac { 22 } { 7 } )$
[Hint: Length of successive semicircles is$ l_1, l_2, l_3, l_4, ...$ with centres at A, B, A, B, ... respectively.]

Answer

According to question we are given that a spiral is made up of successive semi-circles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ....as shown in Fig.
Let $l_1, l_2, l_3, l_4,..l_{13}$ be the lengths (circumferences) of semi-circles of radii $r_1= 0.5 cm, r_2 =1.0 cm, r_3 = 1.5 cm, r_4 = 2.0 cm, r_5 = 2.5 cm,...$ respectively.

Now, Semi-perimeter of circle = $\pi\cdot r $
Therefore,
$l _ { 1 } = \pi r _ { 1 } = \pi \times 0.5 = \frac { \pi } { 2 } \mathrm { cm }$
$l _ { 2 } = \pi r _ { 2 } = \pi \times 1 = 2 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 3 } = \pi r _ { 3 } = \pi \times \frac { 3 } { 2 } = 3 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 4 } = \pi r _ { 4 } = \pi \times 2 = 4 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
and
$l _ { 13 } = \pi r _ { 13 } = \pi \times \frac { 13 } { 2 } \mathrm { cm } = 13 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
Therefore total length of the spiral $= l_1 + l_2 + l_3 +...+ l_{13}$
$\bf= \left\{ \frac { \pi } { 2 } + 2 \left( \frac { \pi } { 2 } \right) + 3 \left( \frac { \pi } { 2 } \right) + \dots + 13 \left( \frac { \pi } { 2 } \right) \right\} $
$\bf= \frac { \pi } { 2 } ( 1 + 2 + 3 + \cdots + 13 ) $
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } ( 1 + 13 ) \quad \left[ \text { Using } S _ { n } = \frac { n } { 2 } ( a + l ) \right]$
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } \times 14$ = $\bf\frac { 1 } { 2 } \times \frac { 22 } { 7 } \times 13 \times 7$ = $\bf {143 cm}$
which is required length of the spiral made up of thirteen consecutive semi-circles.

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Question 65 Marks

A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is ₹ 20 less than its preceding term, find the value of each of the prizes.

Answer

It is given that the sum of seven cash prizes is equal to ₹ 700.
And, each prize is ₹ 20 less than its preceding term.
Let the value of first prize = ₹ a
Let the value of second prize =₹ (a−20)
Let the value of third prize = ₹ (a−40)
So, we have a sequence of the form:
a, a−20, a−40, ...................
It is an arithmetic progression because the difference between consecutive terms is constant.
First term = a, Common difference = d = (a − 20) − a= −20
n = 7 (Because there are total of seven prizes)
$S_7 = ₹ 700$ {given}
Applying formula, ${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$ to find sum of n terms of AP, we get
${S_7} = \frac{7}{2}\left[ {2a + (7 - 1)( - 20)} \right]\;\;$
$\Rightarrow 700 = \frac{7}{2}[2a - 120]$
$\Rightarrow 200 = 2a− 120$
$\Rightarrow 320 = 2a$
$\Rightarrow a =160$
Therefore, value of first prize = ₹ 160
Value of second prize = 160 - 20= ₹ 140
Value of third prize = 140 - 20= ₹ 120
Value of fourth prize = 120 - 20 = ₹ 100
Value of fifth prize = 100 - 20 = ₹ 80
Value of sixth prize = 80 - 20 = ₹ 60
Value of seventh prize = 60 - 20 = ₹ 40

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Question 75 Marks

Find the sum of the odd numbers between 0 and 50.

Answer

The odd numbers between 0 and 50 are 1, 3, 5, 7, ....., 49.
Here, $a_2 - a_4 = 3 - 1 = 2$
$a_3 - a_2 = 5 - 3 = 2$
$a_4 - a_3 = 7 - 5 = 2$
i.e. $a_{k+1} - a_k$ is the same everytime.
So, the above list of numbers forms an AP.
Here, $a = 1$
$d = 2$
$l = 49$
Let the number of terms of the AP be n.
Then, l = a + (n - 1)d
$ \Rightarrow $ 49 = 1 + (n - 1)d
$ \Rightarrow $ (n - 1)2 = 48
$ \Rightarrow n - 1 = \frac{{48}}{2}$
$ \Rightarrow $ n - 1 = 24
$ \Rightarrow $ n = 24 + 1
$ \Rightarrow $ n = 25
Hence, the number of terms of the AP be 25.
$\therefore $ Sum of the odd numbers between 0 and $50 = S_{25}$
$ = \frac{{25}}{2}(a + l)$ ......$\because {S_n} = \frac{n}{2}(a + l)$
$ = \left( {\frac{{25}}{2}} \right)(1 + 49)$
$ = \left( {\frac{{25}}{2}} \right)(50)$
= (25) (25)
= 625

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Question 85 Marks

Show that $a_1, a_2, ........, a_n$, form an AP where $a_n = 9 - 5n.$

Answer

We have $a_n = 9 - 5n$
Put n = 1, 2, 3, 4,.... in succession, we get
$a_1 = 9 - 5(1) = 9 - 5 = 4$
$a_2 = 9 - 5(2) = 9 - 10 = - 1$
$a_3 = 9 - 5(3) = 9 - 15 = -6$
$a_4 = 9 - 5(4) = 9 - 20 = -11$
$\therefore $ $a_2 - a_1 = - 1 - 4 = - 5$
$a_3 - a_2 = -6 - (-1) = -6 + 1 = -5$
$a_4 - a_5 = -11 - (-6) = -11 + 6 = - 5$
$i.e. a_{k+1} - a_k$ is the same everytime
So, $a_1, a_2$, ..., an, ...... form an AP
Here, $a = a_1 = 4$
$a = a_2 - a_1 = -5$
$\therefore $ Sum of the first 15 terms $= S_{15}$
$ = \frac{{15}}{2}\left[ {2a + (n - 1)d} \right]$ $\because {S_n} = \frac{n}{2}[2a + (n - 1)d]$
$ = \frac{{15}}{2}\left[ {2 \times a + (15 - 1)d} \right]$
$ = \frac{{15}}{2}\left[ {2a + 14d} \right]$
$= 15(a + 7d)$
$= (15) [4 + 7(-5)]$
$= (15) (4 - 35)$
$= (15) (-31)$
$= -465$

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Question 95 Marks

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP.

Answer

Let the first term and the common difference of the AP be a and d respectively.
Then,
4th term = a + (4 - 1)d = a + 3d $\because {a_n} = a + (n - 1)d$
8th term = a + (8 - 1)d = a + 7d $\because {a_n} = a + (n - 1)d$
6th term = a + (6 - 1)d = a + 5d $\because {a_n} = a + (n - 1)d$
and 10th term = a + (10 - 1)d = a + 9d $\because {a_n} = a + (n - 1)d$
According to the question,
$4^{th} term + 8^{th} term = 24$
$ \Rightarrow $ (a + 3d) + (a + 7d) = 24
$ \Rightarrow $ 2a + 10d = 24
$ \Rightarrow $ a + 5d = 12 .......... (1) (Dividing throughout by 2)
6th term + 10th term = 24
$ \Rightarrow $ (a + 5d) + (a + 9d) = 44
$ \Rightarrow $ 2a + 14d = 44
$ \Rightarrow $ a + 7d = 22 .......... (2) (Dividing throughout by 2)
Solving (1) and (2), we get
a = -13
d = 15
So, First term = -13
Second term = -13 + 5 = -8
Third term = -8 + 5 = -3
Hence, the first three terms of the given AP are -13, -8 and -3.

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Question 105 Marks

Find the 20th term from the last term of the AP : 3, 8, 13, …., 253.

Answer

The given AP is 3, 8, 13, ..., 253
Here, a = 3
d = 8 - 3 = 5
l = 253
Let the number of terms of the AP be n.
Term, nth term = l
$ \Rightarrow $ 3 + (n - 1)5 = 253 $\because {a_n} = a + (n - 1)d$
$ \Rightarrow $ (n - 1)5 = 253 - 3
$ \Rightarrow $ (n - 1)5 = 250
$ \Rightarrow n - 1 = \frac{{250}}{5}$
$ \Rightarrow $ n - 1 = 50
$ \Rightarrow $ n = 50 + 1
$ \Rightarrow $ n = 51
So, there are 51 terms in the given AP.
Now, 20th term from the last term
= (51 - 20 + 1)th term from the beginning
= 32th term from the beginning
= 3 + (32 - 1)5 $\because {a_n} = a + (n - 1)d$
= 3 + 155
= 158
Hence, the 20th term from the last term of the given AP is 158.
Aliter. Let us write the given AP in the reverse order.
Then the AP becomes 253, 248, 243, ...., 3
Here, a = 253
d = 248 - 253 = -5
Therefore, required term
= 20th term of the AP
= 253 + (20 - 1) (-5) $\because {a_n} = a + (n - 1)d$
= 253 - 95
= 158
Hence, the 20th term from the last term of the given AP is 158.

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Question 115 Marks

Fill in the blanks in the following table, given that a is the first term, $d$ is the common difference and $a_n$ is the $n^{\text {th }}$ term of the AP:

	a	d	n	$a_n$
i	7	3	8	...
ii	-18	...	10	0
iii	...	-3	18	-5
iv	-18.9	2.5	...	3.6
v	3.5	0	105	...

Answer

Given: a = 7, d = 3 and n = 8,
$a_n = ?$
We know that $a_n = a + (n – 1)d$
Thus, $a_8 = 7 + (8 – 1)3 = 7 + 21 = 28$
Given: $a = - 18, n = 10, a_n = 0, d = ?$
We know that $a_n = a + (n – 1)d$
Thus, $0 = - 18 + (10 – 1)d$
$0 = -18 + 9d$
$or, 9d = 18$
$d = 18/9 = 2$
Given: $d = - 3, n = 18, a_n = - 5, a = ?$
We know that $a_n = a + (n – 1)d$
$-5 = a + (18 - 1) (-3)$
$-5 = a - 51$
or a = -5 + 51 = 46
Given: $a = -18.9, d = 2.5, a_n = 3.6, n = ?$
We know that $a_n = a + (n – 1)d$
$3.6 = – 18.9 + (n – 1)2.5$
$or, 2.5(n – 1) = 3.6 + 18.9 = 22.5$
$n – 1 = 22.5/2.5 = 9$
$n = 9 + 1 = 10$
Given: $a = 3.5, d = 0, n = 105, a_n= ?$
We know $a_n = a + (n - 1)d$
$a_{105} = 3.5 + (105 - 1)0$
$a_{105} = 3.5 + 0$
$a_{105} = 3.5$

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