# RS Aggarwal Solutions Class 10 Ex 11B

Q.1: The 24th term of an AP is 5 more than twice its 8th term. Show that its 72nd term is 4 times its 15th term.

Sol:

Let a be the first term and d be the common difference of the AP. Then,

$a_{ 24 } = 2a_{ 10 }$    (Given)

→ a + 23d = 2 ( a + 9d )          $\left [ a_{ n } = a + (n – 1)d \right ]$

→ a + 23d = 2a + 18d

→ 2a – a = 23d – 18d

→ a = 5d              …(1)

Now,

$\frac{ a_{ 72 } }{ a_{ 15 } } = \frac{ a + 71d }{ a +14d }$

$\frac{ a_{ 72 } }{ a_{ 15 } } = \frac{ 5d + 71d }{ 5d + 14d }$    [From (1)]

$\frac{ a_{ 72 } }{ a_{ 15 } } = \frac{ 76d }{ 19d }$ = 4

$a_{ 72 } = 4 \times a_{ 15 }$

Hence, the 72nd term of the AP is 4 times its 15th term.

Q.2: The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.

Sol:

Let a be the first term and d be the common difference of the AP. Then,

$a_{ 19 } = 4 a_{ 6 }$     (Given)

→ a + 18d = 3 ( a + 5d)           $\left [ a_{ n } = a + (n – 1)d \right ]$

→ a + 18d = 3a + 15d

→ 3a – a = 18d – 15d

→ 2a = 3d            …(1)

Also,

$a_{ 9 } = 19$                 (Given)

From (1) and (2), we get

$\frac{ 3d }{ 2 } + 8d = 19$

$\frac{ 3d + 16d }{ 2 } = 19$

→ 19d = 38

→ d = 2

Putting d = 2 in (1), we get

2a = 3 x 2 = 6

→ a = 3

So,

$a_{ 2 }$ = a + d = 3 + 2 = 7

$a_{ 3 }$ = a + 2d = 3 + 2 x 2 = 7,…

Hence, the AP is 3, 5, 7, 9, …

Q.3: If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.

Sol:

In the given AP, let the first term be a and the common difference be d.

Then $T_{ n }$ = a + (n -1) d

$T_{ p }$ = a + (p – 1) d = q          …(i)

$T_{ q }$ = a + (q – 1) d = p          …(ii)

On subtracting (i) from (ii), we get:

(q – p)d = (p – q)

→ d = -1

Putting d = -1 in (i), we get:

a = (q + q – 1)

Thus, a = (p + q – 1) and d = -1

Now, $T_{ p + q }$ = a + (p + q – 1) d

= ( p + q – 1) + ( p + q – 1)( -1)

= ( p + q – 1) – ( p + q – 1) = 0

Hence, the (p + q)th term is 0(zero).

Q.4: The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term from the end is

(a + l).

Sol:

In the given AP, first term = a and the last term = l

Let the common difference be d.

Then, nth term from the beginning is given by

$T_{ n }$ = a + (n – 1) d           …(1)

Similarly, nth term from the end is given by

$T_{ n }$ = {l – (n -1) d }                   …(2)

Adding (1) and (2) , we get:

a + ( n -1 ) d + {l – (n – 1) d }

= a + (n – 1) d + l – (n -1)d

= a + l

Hence, the sum of the nth term from the beginning and the nth term from the end is (a + l)

Q.5: How many two-digit numbers are divisible by 6?

Sol:

The two-digit numbers divisible by 6 are 12, 18, 24, …, 96.

Clearly, these number are in AP

Here, a =12 and d = 18 – 12 = 6

Let this AP contains n terms. Then,

$a_{ n }$ = 96

→ 12 + (n – 1) x 6 = 96            $\left [ a_{ n } = a + (n – 1)d \right ]$

→ 6n + 6 = 96

→ 6n = 96 – 6 = 90

→ n = 15

Hence, there are 15 two-digit numbers divisible by 6.

Q.6: How many two-digit numbers are divisible by 3?

Sol:

The two-digit numbers divisible by 3 are 12, 15, 18, …, 99

Clearly, these numbers are in AP.

Here, a = 12 and d = 15 – 12 = 3

Let this AP contains n terms. Then,

$a_{ n }$ = 99

→ 12 + (n – 1) x 3 = 99            $\left [ a_{ n } = a + (n – 1)d \right ]$

→ 3n + 9 = 99

→ 3n = 99 – 9 = 90

→ n = 30

Hence, there are 30 two-digit numbers divisible by 3.

Q.7: How many three-digit numbers are divisible by 9?

Sol:

The three-digit numbers divisible by 9 are 108, 117, 126, …, 999.

Clearly, these numbers are in AP.

Here, a = 108 and d = 117 – 108 = 9

Let this AP contains n terms. Then,

$a_{ n }$ = 999

→ 108 + ( n – 1 ) x 9 = 999      $\left [ a_{ n } = a + (n – 1)d \right ]$

→ 9n + 99 = 999

→ 9n = 999 – 99 = 900

→ n = 100

Hence, there are 100 three-digit numbers divisible by 9.

Q.8: How many numbers are there between 101 and 999, which is divisible by both 2 and 5?

Sol:

The numbers which are divisible by both 2 and 5 are divisible by 10 also.

Now, the numbers between 101 and 999 which are divisible 10 are 110, 120, 130,…, 990

Clearly, these numbers are in AP.

Here a =110 and s = 120 – 110 = 10

Let this AP contains n terms. Then,

$a_{ n }$ = 990

→ 110 + ( n – 1) x 10 = 990

→ 10n + 100 = 990

→ 10n = 990 – 100 = 890

→ n = 89

Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.

Q.9: In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?

Sol:

The numbers of rose plants in consecutive rows are 43, 41, 39, …, 11

Difference of rose plants between two consecutive rows = (41 – 43) = ( 39 – 41)

= -2 [Constant]

So, the given progression is an AP.

Here, first term = 43

Common difference = -2

Last term = 11

Let n be the last term, then we have:

$T_{ n }$ = a + (n – 1) d

→ 11 = 43 + (n – 1)( -2 )

→ 11 = 45 – 2n

→ 34 = 2n

→ n = 17

Hence, the 17th term is 11 or there are 17 rows in the flower bed.

Q.10: A sum of Rs. 2800 is to be used to award four prizes. If each prize after the first is Rs. 200 less than the preceding prize, find the value of each of the prizes.

Sol:

Let the amount of the first prize be Rs. a

Since each prize after the first is Rs. 200 less than the preceding prize, so the amounts of the four prizes are in AP.

Amount of the second prize = Rs. (a – 200)

Amount of the second prize = Rs. ( a – 2 x 200 ) = Rs. ( a – 400 )

Amount of the fourth prize = Rs. (a – 3 x 200 ) = Rs. (a – 600)

Now,

Total sum of the four prizes = Rs. 2800

Therefore,

Rs. a + Rs. (a – 200) + Rs. (a – 400) + Rs. (a – 600) = Rs. 2800

→ 4a – 1200 = 2800

→ 4a = 2800 + 1200 = 4000

→ a = 1000

Therefore, Amount of the first prize = Rs. 1000

Amount of the second prize = Rs. 1000 – 200 = Rs. 800

Amount of the third prize = Rs. 1000 – 400 = Rs. 600

Amount of the fourth prize = Rs. 1000 – 600 = Rs. 400

Hence, the value of each of the prizes is Rs. 1000, 800, 600, 400.

## Exercise: 11 B

Question 1: Determine K so that (3k -2), (4k – 6) and (k + 2) are three consecutive terms of an A.P.

Sol:

It is given that (3k -2), (4k – 6) and (k + 2) are three consecutive terms of an A.P.

$Therefore, (4k – 6) – (3k -2) = (k + 2) – (4k – 6)$

$4k – 6 – 3k + 2 = k + 2 – 4k + 6$

$\Rightarrow k – 4 = -3k + 8$

$\Rightarrow k + 3k = 4 + 8$

$\Rightarrow 4k = 12$

$\Rightarrow k = 3$

Hence the value of k is 3.

Q.2: Find the value of x for which the numbers (5x + 2), (4x – 1 ) and ( x+ 2) are in A.P.

Sol:

It is given that (5x + 2), (4x – 1 ) and ( x+ 2) are in A.P

$Therefore, (5x + 2) – (4x – 1 ) = (4x – 1) – (x + 2)$

$\Rightarrow 5x + 2 – 4x + 1 = 4x – 1 – x – 2$

$\Rightarrow x + 3 = 3x – 3$

$\Rightarrow 2x = 6$

$\Rightarrow x = 3$

Hence, the values of x is 3.

Q.3: If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an A.P. then find the value of an y.

Sol:

It is given that (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an A.P

$(3y – 1) – (3y + 5) = (3y + 5) – (5y + 1)$

$3y – 1 – 3y – 5 = 3y + 5 – 5y – 1$

$- 6 = 4 – 2y$

$2y = 10$

$y = 5$

Hence the value of y is 5.

Q.4: Find the value of x for which (x+ 2) , 2x, (2x + 3) are three consecutive terms of an A.P.

Sol:

Since are three consecutive terms of an A.P

$(x+ 2) – 2x = 2x – (2x + 3)$

$x+ 2 – 2x = 2x – 2x – 3$

$2 – x = – 3$

$\Rightarrow x = 5$

Hence the value of x is 5.

Q.5: Show that $(a – b)^{2}, (a^{2} + b^{2}) , (a + b)^{2}$ are in A.P.

Sol:

The given numbers are $(a – b)^{2}, (a^{2} + b^{2}) \;\; and \;\; (a + b)^{2}$

For showing the terms given are in AP, we need to prove that

$(a – b)^{2} – (a^{2} + b^{2}) = (a^{2} + b^{2}) – (a + b)^{2}$

$a^{2} + b^{2} – 2ab – a^{2} – b^{2} = a^{2} + b^{2} – a^{2} – b^{2} – 2ab$

$- 2ab = – 2ab$

Thus the common difference is same.

Since each term differs from its preceding term by a constant, therefore the given number are in AP.

Q.6: Find three number in AP whose sum is 15 and product is 80.

Sol:

Let the required numbers be (a – d), a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) x a x (a + d) = a(a2 – d2)

But sum = 15 and product = 80

Therefore,3a = 15$\Rightarrow$a = 5

and a(a2 – d2) = 5 x (25 – d2) = 80 [Since, a = 5]

$\Rightarrow$ (25 – d2) = 16

$\Rightarrow$d2 = 25 – 16$\Rightarrow$ d2 = 9

$\Rightarrow$d = $\pm$3

Thus, a = 5 and d = $\pm$3

Hence, the required numbers are (2, 5, 8) or (8, 5 , 2).

Q.7: The sum of three numbers in AP is 3 and their product is – 35. Find the numbers.

Sol:

Let the required numbers be (a – d), a, (a + d)

Sum of these number = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) x a x (a + d) = a(a2 – d2)

But, sum = 3 and product = – 35

Therefore,3a = 3 $\Rightarrow$ a = 1

and a(a2 – d2) = 1 (1 – d2) = -35

$\Rightarrow$1 – d2 = -35

$\Rightarrow$d2 = 36

$\Rightarrow$d = $\pm$6

Thus, a = 1 and d = $\pm$6

Hence, the required numbers are (-5, 1, 7) or (7, 1, -5)

Let the required number be (a – d), a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) x a x (a – d) = a(a2 – d2)

But sum = 27 and product = 405

Therefore, 3a = 27 a = 9

and a(a2 – d2) = 405

$\Rightarrow$9 x (81 – d2) = 405 [Since,  a = 5]

$\Rightarrow$729 – 9d2 = 405

$\Rightarrow$9d2 = 729 – 405 = 324

$\Rightarrow$d2 = 36

d = ±6

a = 9 and d = 6

Hence the required numbers are (3, 9, 15)

Q.8: Divide 24 in three parts such that they are in AP and their product is 440.

Sol:

Let the required number be (a – d) , a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

Product of these numbers = (a – d) x a x (a + d) = $a(a^{2} – d^{2})$

But sum = 24 and product = 440

Therefore, 3a = 24

$\Rightarrow$ a = 8

and $a(a^{2} – d^{2})$ = 8 (64 – $d^{2})$ ) = 440

$\Rightarrow$64 – $d^{2})$  = 55

$\Rightarrow$ $d^{2})$  = 9

$\Rightarrow$ d = $\pm$3

Thus , a = 8 and d = $\pm$33

Hence the required numbers are (5 , 8 , 11) or (11, 8, 5).

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