5 Marks Questions - MATHS STD 10 Questions - Vidyadip

Questions · Page 1 of 2

5 Marks Questions

🎯

Test yourself on this topic

50 questions · timed · auto-graded

Question 15 Marks

The angles of a quadrilateral are in AP whose common difference is 10º. Find the angles.Hint: Let these angles be xº, (x + 10)º, (x + 20)º and (x + 30)º.
Their sum is 360º.

Answer

Let the required angles be (a - 15)º, (a - 5)º, (a + 5)º, and (a + 15)º as the common difference is 10 (given).
Then (a - 15)º + (a - 5)º + (a + 5)º + (a + 15)º = 360º
⇒ 4a = 360
⇒ a = 90
Hence, the required angles of a quadrilateral are
(90 - 15)º, (90 - 5)º, (90 + 5)º and (90 + 15)º; or 75º, 85º, 95º and 105º.

View full question & answer→

Question 25 Marks

If (2p - 1), 7, 3p are in AP, find the value of p.

Answer

Let (2p - 1), 7 and 3p be three consecutive terms of an AP.
Then 7 - (2p - 1) = 3p - 7
⇒ 5p = 15
⇒ p = 3
$\therefore$ When p = 3, (2p - 1), 7 and 3p form three consecutive terms of an AP.

View full question & answer→

Question 35 Marks

Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.

Answer

Since (x + 2), 2x and (2x + 3) are in AP, we have:
2x - (x + 2) = (2x + 3) - 2x
⇒ x - 2 = 3
⇒ x = 5
$\therefore$ x = 5

View full question & answer→

Question 45 Marks

The sum of first $q$ term of an $A P$ is $\left(63 q-3 q^2\right)$. If its $p^{\text {th }}$ term is -60 , find the value of $p$. Also, find the $11^{\text {th }}$ term of its $A P$.

Answer

Let $S_q$ be the sum of the first m terms of the AP.
$S_q= 63q - 3q^2$
$\Rightarrow S_{q - 1} = 63(q - 1) - 3(q - 1)^2$
$\Rightarrow S_{q - 1} = 63(q - 1) - 3(q - 1)^2$
$\Rightarrow S_{q - 1} = 63q - 63 - 3(q^2 - 2q + 1)$
$\Rightarrow S_{q - 1}= 63q - 63 - 3q^2 + 6q - 3$
$\Rightarrow S_{q - 1 =}-3q^2 + 69q - 66$
Let $a_q$ be the $q^{th}$ term of the AP.
$\therefore a_q=S_q=S_{q-1}$
$\therefore a_q=\left(63 q-3 q^2\right)-\left(-3 m^2+69 q-66\right)$
$\therefore a_q=63 q-3 q^2+3 q^2-69 q+66$
$\therefore a_q=-6 q+66 \ldots . .(i)$
Given that $a_q = -60.$
$\Rightarrow -6p + 66 = -60$
$\Rightarrow -6p = -126$
$\Rightarrow p = 21$
The $11^{\text {th }}$ term $=a_{11}=-6(11)+66=0$.

View full question & answer→

Question 55 Marks

Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.

Answer

It is given that (5x + 2), (4x - 1) and (x + 2) are in AP.
$\therefore$ (4x - 1) - (5x + 2) = (x + 2) = (x + 2) - (4x - 1)
⇒ 4x - 1 - 5x - 2 = x + 2 - 4x + 1
⇒ - x - 3 = -3x + 3
⇒ 3x - x = 3 + 3
⇒ 2x = 6
⇒ x = 3
Hence, the value of x is 3.

View full question & answer→

Question 65 Marks

Find the sum of all natural numbers between 200 and 400 which are divisible by 7.

Answer

Natural numbers between 200 and 400 divisible by 7 are as follows:
203, 210, 217, ....399
Here, $T_n = 399$
$\Rightarrow a + (n - 1)d = 399$
$\Rightarrow 203 + (n - 1)(7) = 399$
$\Rightarrow (n - 1)7 = 196$
$\Rightarrow n - 1 = 28$
$\Rightarrow n = 29$
$\therefore\text{S}_{29}=\frac{29}{2}\big[2\times203+28\times7\big]$
$=\frac{29}{2}\big[406+196]$
$=\frac{29}{2}\times602$
$=8729$

View full question & answer→

Question 75 Marks

What is the $5^{th}$ term from the end of the AP 2, 7, 12, ...47?

Answer

The given AP is 2, 7, 12, ..., 47.
Let us re-write the given AP in reverse order i.e. 47, 42, .... 12, 7, 2.
Now, the $5^{th}$ term from the end of the given AP is equal to the $5^{th}$term from beginning of the AP 47, 42, ...., 12, 7, 2.
Consider the AP 47, 42, ....., 12, 7, 2.
Here, a = 47 and 42 - 47 = -5
$5^{th}$ term of this AP
$= 47 + (5 - 1) \times (-5)$
$= 47 - 20$
$= 27$
Hence, the $5^{th}$ term from the end of the given AP is 27.

View full question & answer→

Question 85 Marks

The first and last terms of an AP are $a$ and I respectively. Show that the sum of the $n ^{\text {th }}$ term from the beginning and the $n ^{\text {th }}$ term from the end is $(a+1)$.

Answer

Let $a$ be the first term and $d$ be the common difference
$\therefore n ^{\text {th }}$ term ftom the beginning $= a +( n -1) d \ldots . .(1)$
$n ^{\text {th }}$ term from end $=1-( n -1) d . .$. (2)
Addind (1) and (2),
Sum of the $n ^{\text {th }}$ term from the beginning and $n^{\text {th }}$ term from the end $=[a+(n-1) d]+[1-(n-1) d]=a+1$

View full question & answer→

Question 95 Marks

There are 25 trees at equal distances of 5m in a line with a water tank, the distance of the water tank from the nearest tree being 10m A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next. Find the total distance covered by the gardener in order to water all the trees.

Answer

Assume that gardener is standing near the well initially and he did not return to the well after watering the last tree. Distance covered by gardener to water $1^{\text {st }}$ tree and return to the initial position
$=10 m+10 m=20 m$
Distance covered by gardener to water $2^{\text {nd }}$ tree and return to the initial position
$=15 m+15 m=30 m$
Distance covered by gardener to water $3^{\text {rd }}$ tree and return to the initial position
$=20 m+20 m=40 m$
$\therefore$ Distances covered by the gardener to water the plants are in AP.
Here $a=20, d=10$
Total distance covered the gardener is given by Sn , where $n =25$.
$\Rightarrow S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{25}=\frac{25}{2}[2(20)+(25-1) 10]=3500$
Thus, the total distance covered by the gardener is 3500 m .

View full question & answer→

Question 105 Marks

Determine k so that (3k - 2), (4k - 6) and (k + 2) are three consecutive terms of an AP.

Answer

It is given that (3k - 2), (4k - 6) and (k + 2) are three consecutive terms of an AP.
$\therefore$ (4k - 6) - (3k - 2) = (k + 2) - (4k - 6)
⇒ 4k - 6 - 3k + 2 = k + 2 - 4k + 6
⇒ k - 4 = -3k + 8
⇒ k + 3k = 8 + 4
⇒ 4k = 12
⇒ k = 3
Hence, the value of k is 3.

View full question & answer→

Question 115 Marks

The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.

Answer

Let the required terms be (a - d), a and (a + d).
Then $(a - d) + a + (a + d) = 21$
$\Rightarrow 3a = 21$
$\Rightarrow a = 7$
Also, $(a - d)^2 + a^2 + (a + d)^2 = 165$
$\Rightarrow 3a^2 + 2d^2 = 165$
$\Rightarrow (3 \times 49 + 2d^2) = 165$
$\Rightarrow 2d^2 = 165 - 147 = 18$
$\Rightarrow d^2 = 9$
$\Rightarrow\text{d}=\pm3$
Thus, a = 7 and $\text{d}=\pm3$
Hence, the required numbers are (4, 7, 10) or (10, 7, 4).

View full question & answer→

Question 125 Marks

How many three-digit numbers are divisible by 9?

Answer

The two-digit numbers divisible by 9 start from
108, 117, 126, 135, ..., 999
Here,
$a = 108$
$d = 9$
$a_n = a + (n - 1)d$
$\Rightarrow 999 = 108 + (n - 1)(9)$
$\Rightarrow 999 = 108 + 9n - 9$
$\Rightarrow 900 = 9n$
$\Rightarrow n = 100$
This, 100 two-digit number are divisible by 9.

View full question & answer→

Question 135 Marks

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

Answer

Let the required parts of 24 be (a - d), a and (a + d) such that they are in AP.
Then $(a - d) + a + (a + d) = 24$
$\Rightarrow 3a = 24$
$\Rightarrow a = 8$
Also, $(a - d) \times a \times (a + d) = 440$
$\Rightarrow a(a^2 - d^2) = 440$
$\Rightarrow 8(64 - d^2) = 440$
$\Rightarrow d^2 = 64 - 55 = 9$
$\Rightarrow\text{d}=\pm3$
Thus, a = 8 and $\text{d}=\pm3$
Hence, the required parts of 24 are (5 , 8, 11) or (11, 8, 5).

View full question & answer→

Question 145 Marks

Find the sum of all three-digit natural numbers which are divisible by 13.

Answer

All three-digit numbers which are divisible by 13 are 104, 117, 130, 143, .…… 938.
This is an AP in which a = 104, d = (117 – 104) = 13 and l = 938
Let the number of terms be n
Then $T_n = 938$
$\Rightarrow a + (n - 1)d = 988$
$\Rightarrow 104 + (n - 1) \times 13 = 988$
$\Rightarrow 13n = 897$
$\Rightarrow n = 69$
$\therefore\text{Required sum}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$=\frac{69}{2}\big[104+988\big]=69\times546=37674 $
Hence, the required sum is 37674.

View full question & answer→

Question 155 Marks

Find three numbers in AP whose sum is 15 and product is 80.
Hint: Let the numbers be (a - d), a, (a + d).

Answer

Let the required numbers be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 15
⇒ 3a = 15
⇒ a = 5
Also, $(a - d) \times a \times (a + d) = 80$
$\Rightarrow a(a^2 - d^2) = 80$
$\Rightarrow 5(25 - d^2) = 80$
$\Rightarrow d^2 = 25 - 16 = 9$
$\Rightarrow\text{d}=\pm3$
Thus, a = 5 and $\text{d}=\pm3$
Hence, the required numbers are (2, 5 and 8) or (8, 5 and 2).

View full question & answer→

Question 165 Marks

Find how many intergers between 200 and 500 are divisible by 8.

Answer

The first term between 200 and 500 divisible by 8 is 208, and the last term is 496.
So, first term (a) = 208
Common difference $(d) = 8$
$a_n= a + (n - 1)d = 496$
$\Rightarrow 208 + (n - 1)8 = 496$
$\Rightarrow (n - 1)8 = 288$
$\Rightarrow n - 1 = 36 \Rightarrow n = 37$
Hence, there are 37 integers between 200 and 500 which are divisible by 8.

View full question & answer→

Question 175 Marks

The first three terms of an AP are respectively (3y - 1), (3y + 5) and (5y + 1), find the value of y.

Answer

The terms (3y - 1), (3y + 5) and (5y + 1) are in AP.
$\therefore$ (3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)
⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5
⇒ 6 = 2y - 4
⇒ 2y = 10
⇒ y = 5
Hence, the value of y is 5.

View full question & answer→

Question 185 Marks

Find the sum of first n natural numbers.

Answer

The first n natural numbers are 1, 2, 3, 4, 5, ..., n.
Here, a = 1 and d = (2 - 1) = 1
Sum of n terms of an AP is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times1+(\text{n}-1)\times1\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2+\text{n}-1\big]=\big(\frac{\text{n}}{2}\big)\times(\text{n}+1)=\frac{\text{n}(\text{n}+1)}{2}$

View full question & answer→

Question 195 Marks

The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first and the common difference of the AP.

Answer

Let a be the first term and d be the common difference of the given A.P.
Then,
$\text{S}_9=81$
$\Rightarrow\frac{9}{2}\big[2\text{a}+8\text{d}\big]=81$
$\Rightarrow\frac{9\times2}{2}\big[\text{a}+4\text{d}\big]=81$
$\Rightarrow\text{a}+4\text{d}=9\dots(\text{i})$
Also, $\text{S}_{20}=400$
$\Rightarrow\frac{20}{2}\big[2\text{a}+19\text{d}\big]=400$
$\Rightarrow10\big[2\text{a}+19\text{d}\big]=400$
$\Rightarrow2\text{a}+19\text{d}=40\dots(\text{ii})$
Multiplying equation (i) by 2, we get
$2\text{a}+8\text{d}=18\dots(\text{iii})$
Subtracting (iii) from (ii), we get
$11\text{d}=22$
$\Rightarrow\text{d}=2$
$\Rightarrow\text{a}=9-4(2)=9-8=1$
Thus, the first term is 1 and the common difference is 2.

View full question & answer→

Question 205 Marks

$Find the sum of the following arithmetic series:$
$34 + 32 + 30 + .... + 10.$

Answer

we havea = 34,
d = 32 - 34 = -2
Let the total number of terms be n.
Then$ T_n = 10$
$⇒ \text{a} + (\text{n} - 1)\text{d} = 10$
$\Rightarrow34+(\text{n}-1)(-2)=10$
$\Rightarrow(\text{n}-1)(-2)=-24$
$\Rightarrow\text{n}-1=\frac{-24}{-2}$
$\Rightarrow\text{n}-1=12$
$\Rightarrow\text{n}=13$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\therefore\text{S}_\text{13}=\frac{\text{13}}{2}\big[\text{34}+10\big]$
$=\frac{13}{2}\times44=286$

View full question & answer→

Question 215 Marks

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 flower bed?

Answer

Number of rose plants in first, second, third rows ....are 43, 41, 39 .... respectively.
There are 11 rose plants in the last row
So, it is an AP. viz. 43, 41, 39 ....11
a = 43, d = 41 - 43 = -2, l = 11
Let $n^{th}$ term be the last term
$\therefore$$ l = a + (n - 1)d$
$\Rightarrow 11 = 43 + (n - 1) \times (n - 1) \times (-2)$
$43 - 2n + 2 = 11 or 2n = 45 - 11 = 34$
$\therefore\text{n}=\frac{34}{2}=17$
Hence, there are 17 rows in the flower bed.

View full question & answer→

Question 225 Marks

If the numbers (2n - 1), (3n + 2) and (6n - 1) are in AP, find the value of n and the numbers.

Answer

It is given that the numbers (2n - 1), (3n + 2) and (6n - 1) are in AP.
$\therefore$ (3n + 2) - (2n - 1) = (6n - 1) - (3n + 2)
⇒ 3n + 2 - 2n + 1 = 6n - 1 - 3n - 2
⇒ n + 3 = 3n - 3
⇒ 2n = 6
⇒ n = 3
When n = 3
2n - 1 = 2 × 3 - 1 = 6 - 1 = 5
3n + 2 = 3 × 3 + 2 = 9 + 2 = 11
6n - 1 = 6 × 3 - 1 = 18 - 1 = 17
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.

View full question & answer→

Question 235 Marks

Find an AP whose $4^{\text {th }}$ term is 9 and the sum of its $6^{\text {th }}$ and $13^{\text {th }}$ terms is 40 .

Answer

Let a be the term and d be the common difference of the AP. Then,
$a_4= 9$
$\Rightarrow a + (4 - 1)d = 9 [a_n = a + (n - 1)d]$
$\Rightarrow a + 3d = 9 ....(1)$
Now,
$a_6+ a_{13} = 40 (given)$
$\Rightarrow (a + 5d) + (a + 12d) = 40$
$\Rightarrow 2a + 17d = 40 ....(2)$
From (1) and (2), we get
$2(9 - 3d) + 17d = 40$
$\Rightarrow 18 - 6d + 17d = 40$
$\Rightarrow 11d = 40 - 18 = 22$
$\Rightarrow d = 2$
Putting d = 2 in (1), we get
a + 3 × 2 = 9
⇒ a = 9 - 6 = 3
Hence, the AP is 3, 5, 9, 11, .....

View full question & answer→

Question 245 Marks

What is the common difference of an AP in which $a_{27} - a_7 = 84?$

Answer

Let 'a' is the first term and d is the common difference of the AP
Given:
$a_{27} - a_7 = 84$
$a_n= a + (n - 1)d$
$a_{27}= a + (27 - 1)d$
$a_7= a + (7 - 1)d$
$a_{27 -} a_7= 84$
$a + 26d - (a + 6d) = 84$
$a + 26d - a - 6d = 84$
$a - a + 26d - 6d = 84$
$26d - 6d = 84$
$20d = 84$
$\text{d} = \frac{84}{20} = \frac{21}{5}$
d = 4.2

View full question & answer→

Question 255 Marks

Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.

Answer

Their sum
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 28
4a = 28
a = 7.
Their product
$(a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 216$
$4a^2+ 20d^2 = 216$
$a^2 + 5d^2 = 54$
Putting the value of a
$49 + 5d^2 = 54$
$5d^2 = 5$
$d^2 = 1$
$\text{d}=\pm1.$
Hence, the numbers are 4, 6, 8 & 10.

View full question & answer→

Question 265 Marks

The sum of the first $n$ terms of an $A P$ is $\left(3 n^2+6 n\right)$. Find the $n^{\text {th }}$ term and the $15^{\text {th }}$ term of this $A P$.

Answer

we have, $S_n = 3n^2 + 6n$
$\Rightarrow S_{n - 1} = 3(n - 1)^2 + 6(n - 1)$
$= 3(n^2 - 2n + 1) + 6n - 6$
$\Rightarrow 3n^2 - 6n + 3 + 6n - 6$
$= 3n^2 - 3$
Now, $n^{th}$term, $T_n = S_n - S_{n - 1}$
$= 3n^2 + 6n - 3n^2 + 3 = 6n + 3$
And, $15^{th}$ term $= T_{15} = 6(15) + 3$
= 90 + 3 = 93

View full question & answer→

Question 275 Marks

Is -150 a term of the AP 11, 8, 5, 2, ....?

Answer

The given AP is 11, 8, 5, 2, ...,
Common difference = 8 - 11 = -3
The given term is -150.
The general term of an AP is given by
$a_n = a + (n - 1)d$
$\Rightarrow -150 = a + (n - 1)d$
$\Rightarrow -150 = 11 + (n - 1)(-3)$
$\Rightarrow -150 = 11 - 3n + 3$
$\Rightarrow 3n - 3 = 161$
$\Rightarrow 3n = 164$
$\Rightarrow\text{n}=\frac{164}{3}$
The number of terms cannot be a fraction.
So clearly, -150 is not a term of the AP.

View full question & answer→

Question 285 Marks

The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

Let theire be n terms in the given A.P.
First terms, a = 17
Last term, l = 350
Common defference, d = 9
Now, $T_n = 350$
$\Rightarrow a + (n - 1)d = 350$
$\Rightarrow 17 + (n -1)9 = 350$
$\Rightarrow (n - 1)9 = 333$
$\Rightarrow n - 1 = 37$
$\Rightarrow n = 38$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\Rightarrow\text{S}_\text{38}=\frac{\text{38}}{2}\big[\text{17}+\text{350}\big]=19\times367=6973$

View full question & answer→

Question 295 Marks

Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its $25^{th}$ term.

Answer

$\text{S}_{14}=1505$ (Given)
$\Rightarrow\frac{14}{2}\big[2\times10+(14-\text{l})\times\text{d}\big]=1505$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow 7(20 + 13d) = 1505$
$\Rightarrow 20 + 13d = 215$
$\Rightarrow 13d = 215 - 20 = 195$
$\Rightarrow d = 15$
$\therefore$ $25^{th}$ term of the AP, $a_{25}$
$= 10 + (25 - l) \times 15$
$[a_n = a + (n - l)d]$
$= 10 + 360$
$= 370$
Hence the required term is 370.

View full question & answer→

Question 305 Marks

The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.
Hint: Let these terms be (a - b), a, (a + d).

Answer

Let the first three terms of the AP be (a - d), a and (a + d),
Then,
(a - d) + a + (a + d) = 48
⇒ 3a = 48
⇒ a = 16
Now,
(a - d) × a = 4(a + d) + 12
⇒ (16 - d) × 16 = 4 (16 + d) + 12
⇒ 256 - 16d = 64 + 4d + 12
⇒ 16d + 4d = 256 - 76
⇒ 20d = 180
⇒ d = 9
When a = 16 and d = 9,
a - d = 16 - 9 = 7
a + d = 16 + 9 = 25
Hence, the first three terms of the AP are 7, 16 and 25.

View full question & answer→

Question 315 Marks

The first term of an AP is p and its common difference is q. Find its $10^{th}$ term.

Answer

Here, a = p and d = q
Now, $T_n = a + (n - 1)d$
$\Rightarrow T_n = p + (n - 1)q$
$\therefore$ $T_{10}= p + 9q$

View full question & answer→

Question 325 Marks

Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and third terms as 7 : 15
Hint: Let these parts be (a - 3d), (a - d), (a + d) and (a + 3d).

Answer

Let the four parts in AP be (a - 3d), (a - d), (a + d) and (a + 3d).
Then,
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = 8 ....(1)
Also,
(a - 3d)(a + 3d) : (a - d)(a + d) = 7 : 15
$\Rightarrow\frac{(8-3\text{d})(8+3\text{d})}{(8-\text{d})(8+\text{d})}=\frac{7}{15}$ [From (1)]
$\Rightarrow\frac{64-9\text{d}^2}{64-\text{d}^2}=\frac{7}{15}$
$\Rightarrow15(64-9\text{d}^2)=7(64-\text{d}^2)$
$\Rightarrow960-135\text{d}^2=448-7\text{d}^2$
$\Rightarrow135\text{d}^2-7\text{d}^2=960-448$
$\Rightarrow128\text{d}^2=512$
$\Rightarrow\text{d}^2=4$
$\Rightarrow\text{d}=\pm2$
When a = 8 and d = 2,
$\text{a}-3\text{d}=8-3\times2=8-6=2$
$\text{a}-\text{d}=8-2=6$
$\text{a}+\text{d}=8+2=10$
$\text{a}+3\text{d}=8+3\times2=8+6=14$
When a = 8 and d = -2,
$\text{a}-3\text{d}=8-3\times(-2)=8+6=14$
$\text{a}-\text{d}=8-(-2)=8+2=10$
$\text{a}+\text{d}=8-2=6$
$\text{a}+3\text{d}=8+3\times(-2)=8-6=2$
Hence, the four parts are 2, 6, 10 and 14.

View full question & answer→

Question 335 Marks

What is the sum of first n terms of the AP a, 3a, 5a, ....

Answer

The given AP is a, 3a, 5a, ...
Here
First term, A = a
Common difference, D = 3a - a = 2a
$\therefore$ sum of first n terms, $S_n$
$=\frac{\text{n}}{2}\big[2\times\text{a} + (\text{n}-1)\times2\text{a}\big]$ $\Big\{\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{A} + (\text{n}-1)\text{D}\big]\Big\}$
$=\frac{\text{n}}{2}(2\text{a}+2\text{an}-2\text{a})$
$=\frac{\text{n}}{2}\times2\text{an}$
$=\text{an}^2$
Hence, the required sum is $an^2$.

View full question & answer→

Question 345 Marks

If 10 times the $10^{\text {th }}$ term of an AP is equal to 15 times the $15^{\text {th }}$ term, show that its $25^{\text {th }}$ term is zero.

Answer

Let a be the first term and d be the common difference
$T_{10} = a + 9d, T_{15}= a + 14d$
$10T_{10} = 15T_{15}$
$\Rightarrow 10 (a + 9d) = 15(a + 14d)$
$or 2(a + 9d) = 3(a + 14d)$
$\Rightarrow a + 24d = 0$
$\therefore$ $T_{25} = 0$

View full question & answer→

Question 355 Marks

The sum of three numbers in AP is 3 and their product is -35. Find the numbers.

Answer

Let the required numbers be (a - d), a and (a + d).
Then $(a - d) + a + (a + d) = 3$
$\Rightarrow 3a = 3$
$\Rightarrow a = 1$
Also, $(a - d) \times a \times (a + d) = -35$
$\Rightarrow a(a^2 - d^2) = -35$
$\Rightarrow 1 \times (1 - d^2) = -35$
$\Rightarrow d^2 = 36$
$\Rightarrow\text{d}=\pm6$
Thus, a = 1 and $\text{d}=\pm6$
Hence, the required numbers are (-5, 1 and 7) or (7, 1 and -5).

View full question & answer→

Question 365 Marks

Which term of the AP 21, 18, 15, .... is zero?

Answer

In the given AP, first term, a = 21 and common difference, d = (18 - 21) = -3
Let's its $n^{th} $term be 0.
Then $T_n = 0$
$\Rightarrow a + (n - 1)d = 0$
$\Rightarrow 21 + (n - 1) \times (-3) = 0$
$\Rightarrow 24 - 3n = 0$
$\Rightarrow 3n = 24$
$\Rightarrow n = 8$
Hence, the $8^{th}$ term of the given AP is 0.

View full question & answer→

Question 375 Marks

Write the next term of the $\text{AP}\sqrt{8},\sqrt{18},\sqrt{32},....$

Answer

The given AP is $\sqrt{8},\sqrt{18},\sqrt{32},....$
On simplifying the terms, we get :
$2\sqrt{2},3\sqrt{2},4\sqrt{2},....$
Here, $\text{a}=2\sqrt{2}$ and $\text{d}=(3\sqrt{2}-2\sqrt{2})=\sqrt{2}$
$\therefore$ Next term, $\text{T}_4=\text{a}+3\text{d}=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}=\sqrt{50}$

View full question & answer→

Question 385 Marks

The sum of first $m$ term of an $A P$ is $\left(4 m^2-m\right)$. If its $n^{\text {th }}$ term is 107 , find the value of $n$. Also, find the $21^{\text {st }}$ term of this $A P$.

Answer

Let $S _{ m }$ be the sum of the first m terms of the AP.
$S_m=4 m^2-m$
$\Rightarrow S_{m-1}=4(m-1)^2-(m-1)$
$\Rightarrow S_{m-1}=4\left(m^2-2 m+1\right)-m+1$
$\Rightarrow S_{m-1}=4 m^2-8 m+4-m+1$
$\Rightarrow S_{m-1}=4 m^2-9 m+5$
Let $a_m$ be the $m^{\text {th }}$ term of the AP.
$\therefore a_m=S_m=S_{m-1}$
$\therefore a_m=\left(4 m^2-m\right)-\left(4 m^2-9 m+5\right)$
$\therefore a_m=4 m^2-m-4 m^2+9 m-5$
$\therefore a_m=8 m-5 \ldots . . \text { (i) }$
Given that $a_n=107$.
$\Rightarrow 8 n-5=107$
$\Rightarrow 8 n=112$
$\Rightarrow n=14$
The $21^{\text {st }}$ term $=a_{21}=8(21)-5=163$

View full question & answer→

Question 395 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 precending prize, find the value of each prize.

Answer

Let the $1^{\text {st }}$ prize be $a$
So the $2^{\text {nd }}$ prize $=a-20$
The $3^{\text {rd }}$ prize $=(a-20-20=a-40$
And so on
So, the series will be
$a, a-20, a-40, \ldots$
Since difference is same, it is an AP
Here, sum of Rs 700 is given,
$\text { So, } S_n=700$
Also,
$a=a$
$d=(a-20)-a=a-20-a=-20$
We know that
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
Putting the values
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$700=\frac{7}{2}(2\text{a}+(7-1)\times-20)$
$700=\frac{7}{2}(2\text{a}+6\times-20)$
$700=\frac{7}{2}(2\text{a}-120)$
$700\times\frac{7}{2}=2\text{a}-120$
$200=2\text{a}-120$
$200+120=2\text{a}$
$320=2\text{a}$
$\frac{320}{2}=\text{a}$
$160=\text{a}$
$\text{a}=160$
So,
$1^{st} prize = a = 160$
$2^{nd} prize = a - 20 = 160 - 20 = 140$
$3^{rd}prieze = a - 40 = 160 - 40 = 120$
$4^{th}prize = a - 60 = 160 - 60 =100$
$5^{th} prize = a - 80 = 160 - 80 = 80$
$6^{th} prize = a - 100 = 160 - 100 = 60$
$7^{th} prize = a - 120 = 160 - 120 = 40$

View full question & answer→

Question 405 Marks

Show that $(a-b)^2,\left(a^2+b^2\right)$ and $(a+b)^2$ are in AP.

Answer

The given number are $(a - b)^2, (a^2 + b^2)$ and $(a + b)^2$.
Now,
$(a^2 + b^2) - (a - b)^2 = a^2 + b^2 - (a^2 - 2ab + b^2) = a^2 + b^2 - a^2 + 2ab - b^2 = 2ab$
$(a + b)^2 - (a^2+ b^2) = a^2 + 2ab + b^2 - a^2 - b^2 = 2ab$
$So, (a^2 + b^2) - (a - b)^2 = (a + b)^2 - (a^2 + b^2) = 2ab$ (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.

View full question & answer→

Question 415 Marks

How many terms of the AP 21, 18, 15, ..... must be added to get the sum 0?

Answer

Here a = 21, d = (18 - 21) = -3
Let the required number of terms be n, then
$\text{S}_\text{n}=0\Rightarrow\frac{\text{}\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]=0$
$\Rightarrow\frac{\text{n}}{2}\big[2\times21+(\text{n}-1)(-3)\big]=0$
$\Rightarrow\frac{\text{n}}{2}(45-3\text{n})=0$
$\Rightarrow\text{n}(45-3\text{n})=0$
$\Rightarrow45-3\text{n}=0\Rightarrow3\text{n}=45$
$\Rightarrow\text{n}=15$
$\therefore$ Sum of first 15 terms = 0

View full question & answer→

Question 425 Marks

The sum of the first n terms of an AP is $\Big(\frac{3\text{n}^2}{2}+\frac{5\text{n}}{2}\Big).$ Find its $n^{th}$ term and the $25^{th}$ term.

Answer

we have, $\text{S}_\text{n}=\frac{3\text{n}^2}{2}+\frac{5\text{n}}{2}=\frac{3\text{n}^2+5\text{n}}{2}$
$\Rightarrow\text{S}_{\text{n}-1}=\frac{3(\text{n}-1)^2+5(\text{n}-1)}{2}$
$=\frac{3(\text{n}^2-2\text{n}+1)+5\text{n}-5}{2}$
$=\frac{3\text{n}^2-6\text{n}+3+5\text{n}-5}{2}$
$=\frac{3\text{n}^2-\text{n}-2}{2}$
Now, $n^{th}$ term
$ \text{T}_\text{n} = \text{S}_\text{n} - \text{S}_{\text{n} - 1}$
$=\frac{3\text{n}^2+5\text{n}}{2} - \frac{3\text{n}^2-\text{n}-2}{2}$
$=\frac{3\text{n}^2+5\text{n}-3\text{n}^2+\text{n}+2}{2}$
$=\frac{6\text{n}+2}{2}=3\text{n}+1$
$25^{th}$ term $= T_{25} = 3(25) + 1 = 75 + 1 = 76$

View full question & answer→

Question 435 Marks

The sum of the first 7 terms of an $A P$ is 182 . If its $4^{\text {th }}$ and $17^{\text {th }}$ terms are in the ratio $1: 5$, find the $A P$.

Answer

Let a be the first term and d be the common difference of the given A.P.
Then,
$T_4 = a + 3d and T_{17} = a + 16d$
Now, $\frac{\text{T}_4}{\text{T}_{17}}=\frac{1}{5}$
$\Rightarrow\frac{\text{a}+3\text{d}}{\text{a}+16\text{d}}=\frac{1}{5}$
$\Rightarrow5\text{a}+15\text{d}=\text{a}+16\text{d}$
$\Rightarrow4\text{a}-\text{d}=0$
$\Rightarrow4\text{a}=\text{d}\dots(\text{i})$
Also, $\text{S}_7=182$
$\Rightarrow\frac{7}{2}\big[2\text{a}+6\text{d}\big]=182$
$\Rightarrow\frac{7\times2}{2}\big[\text{a}+3\text{d}\big]=182$
$\Rightarrow\text{a}+3\text{d}=26$
$\Rightarrow\text{a}+3\text{(4a})=26\dots\big[\text{From(i)}\big]$
$\Rightarrow13\text{a}=26$
$\Rightarrow\text{a}=2$
$\Rightarrow\text{d}=4(2)=8$
Thus, we have
$T_1 = 2$
$T_2= T_1 + d = 2 + 8 = 10$
$T_3= T_1 + 2d = 2 + 2(8) = 2 + 16 = 18$
$T_4 = T_1 + 3d = 2 + 3(8) = 2 + 24 = 26$
Thus, the given A.P. is 2, 10, 18, 26, .....

View full question & answer→

Question 445 Marks

In an AP, the first term is 2, the last term is 29 and the sum of all the terms is 155. Find the common difference.

Answer

First term 'a' of an AP = 2
The last term l = 29
Sum of n term $\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})=155$
$\therefore\frac{\text{n}}{2}(\text{2}+29)=155$
$\text{n}=\frac{155\times2}{31}=10$
Also, l = a + (n - 1)d
or 29 = 2 + (10 - 2)d = 2 + 9d
⇒ 9d = 29 - 2 = 27 $\therefore\text{d}=\frac{27}{9}=3$
$\therefore$ common difference = 3

View full question & answer→

Question 455 Marks

If (3y - 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.

Answer

It is given that (3y - 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.
$\therefore$ (3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)
⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5
⇒ 6 = 2y - 4
⇒ 2y = 6 + 4 = 10
⇒ y = 5
Hence, the value of y is 5.

View full question & answer→

Question 465 Marks

How many numbers are there between 101 and 999, which are divisible by both 2 and 5?

Answer

For the number to be divisible by both 2 and 5, they have to be divisible by the LCM of 2 and 5 = 10.
The numbers divisible by 10 between 101 and 999
Are 110, 120, 130, ...., 990
Here
$a = 110$
$d = 10$
$a_n = a + (n - 1)d$
$\Rightarrow 990 = 110 + (n - 1)(10)$
$\Rightarrow 990 = 110 + 10n - 10$
$\Rightarrow 890 = 10n$
$\Rightarrow n = 89$
Thus, 89 numbers between 101 and 999 are divisible by both 2 and 5.

View full question & answer→

Question 475 Marks

How many two-digit numbers are divisible by 6?

Answer

The two-digit numbers divisible by 6 start from
12, 18, 24, ..., 96
Here,
$a = 12$
$d = 6$
$a_n = a + (n - 1)d$
$\Rightarrow 96 = 12 + (n - 1)(6)$
$\Rightarrow 96 = 12 + 6n - 6$
$\Rightarrow 90 = 6n$
$\Rightarrow n = 15$
This, 15 two-digit number are divisible by 6.

View full question & answer→

Question 485 Marks

The $12^{th}$ term of an AP is -13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.

Answer

$12^{th} term = T_{12} = -13$
$\Rightarrow a + 11d = -13 ....(i)$
$S_4 = 24$
$\frac{4}{2}\big[2\text{a}+3\text{d}\big]=24$
$\Rightarrow2\big[2\text{a}+3\text{d}\big]=24$
$\Rightarrow2\text{a}+3\text{d}=12\dots(\text{ii})$
Multiplying equation (i) by 2, we get
2a + 22d = -26 .....(iii)
Subtrating (iii) from (ii), we get
19d = -38
$\Rightarrow d = -2$
$\Rightarrow a + 11(-2) = -13 ....[From(i)]$
$\Rightarrow a - 22 = -13$
$\Rightarrow a = 9$
$\therefore$ Sum of first 10 terms,
$\text{S}_{10}=\frac{10}{2}\big[2(9)+9(-2)\big]$
$=5[18-18]=5\times0=0$

View full question & answer→

Question 495 Marks

If 18, a, (b - 3) are in AP, then find the value of (2a - b).

Answer

It is given that 18, a, (b - 3) are in AP.
$\therefore$ a - 18 = (b - 3) - a
⇒ a + a - b = 18 - 3
⇒ 2a - b = 15
Hence, the required value is 15.

View full question & answer→

Question 505 Marks

If the sum of first m terms of an AP is $(2m^2+ 3m)$ then what is its second term?

Answer

Let $S_m$ denotes the sum of first $m$ terms of the $A P$.
$\therefore S_m=2 m^2+3 m$
$\Rightarrow S_{m-1}=2(m-1)^2+3(m-1)=2\left(m^2-2 m+1\right)+3(m-1)=2 m^2-m-1$
Now,
$m^{\text {th }} \text { term of the AP, } a_m=S_m-S_{m-1}$
$\therefore a_m=\left(2 m^2+3 m\right)-\left(2 m^2-m-1\right)=4 m+1$
Putting $m=2$, we get
$a_2=4 \times 2+1=9$
hence, the second term of the $A P$ is 9 .

View full question & answer→

Generate Paper Free All Arithmetic Progressions topics