Class 10 Exam > Class 10 Notes > RD Sharma Solutions for Class 10 Mathematics > Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-4)
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-4) | RD Sharma Solutions for Class 10 Mathematics PDF Download
Page No 5.25
Ques.10. If 9th term of an A.P is zero, prove that its 29th term is double the 19th term.
Ans. In the given problem, the 9th term of an A.P. is zero.
Here, let us take the first term of the A.P as a and the common difference as d
So, as we know,
an = a + (n - 1)d
We get,
a9 = a + (9 - 1)d
0 = a + 8 d
a = -8d ...(1)
Now, we need to prove that 29th term is double of 19th term. So, let us first find the two terms.
For 19th term (n = 19),
a19 = a + (19 - 1)d
= -8d + 18d (Using 1)
= 10d
For 29th term (n = 29),
a29 = a + (29 − 1)d (Using 1)
= −8d + 28d
= 20d
= 2 × 10d
=2 × a19
Therefore, for the given A.P. the 29th term is double of the 19th term.
Hence proved.
Ques.11. If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.
Ans. Here, let us take the first term of the A.P. as a and the common difference as d
We are given that 10 times the 10th term is equal to 15 times the 15th term. We need to show that 25th term is zero.
So, let us first find the two terms.
So, as we know,
an = a + (n - 1)d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
For 15th term (n = 15),
a15 = a + (15 - 1)d
= a + 14d
Now, we are given,
10(a + 9d) = 15(a + 14d)
Solving this, we get,
10a + 90d = 15a + 210d
90d - 210d = 15a - 10a
- 120d = 5a
-24d = a ...(i)
Next, we need to prove that the 25th term of the A.P. is zero. For that, let us find the 25th term using n = 25,
a25 = a + (25 - 1)d
= -24d + 24d (Using 1)
= 0
Thus, the 25th term of the given A.P. is zero.
Hence proved
Ques.12. In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Ans. Here, we are given that 24th term is twice the 10th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d
We have to prove that a72 = 2a34
So, let us first find the two terms.
As we know,
an = a + (n - 1)d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
For 24th term (n = 24),
a24 = a + (24 - 1)d
= a + 23d
Now, we are given that a24 = 2a10
So, we get,
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
23d - 18d = 2a - a
5d = a ...(1)
Further, we need to prove that the 72nd term is twice of 34th term. So let now find these two terms,
For 34th term (n = 34),
a34 = a + (34 - 1)d
= 5d + 33d (Using 1)
= 38d
For 72nd term (n = 72),
a72 = a + (72 - 1)d
= 5d + 71d (Using 1)
= 76d
= 2(38d)
Therefore, a72 = 2a34
Ques.13. The 26th, 11th and last term of an A.P. are 0, 3 and -(1/5), respectively. Find the common difference and the number of terms.
Ans. It is given that, a26 = 0, a11 = 3, an = −(1/5).
⇒ a + 25d = 0 .....(i)
⇒ a + 10d = 3 .....(ii)
Subtracting (ii) from (i).
15d = −3
⇒ d = −(1/5)
⇒ a = −25d = 5
Now, an = a +(n − 1)d
⇒ −(1/5) = 5 + (n − 1)(−(1/5))
⇒ −(1/5) - 5 = (n−1)(−(1/5))
⇒ n = 27
Ques.14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
Ans. In the given problem, let us take the first term as a and the common difference as d
Here, we are given that,
a4 = 3a ...(1)
a7 = 2a3 + 1 ...(2)
We need to find a and d
So, as we know,
an = a + (n - 1)d
For the 4th term (n = 4),
a4 = a + (4 - 1)d
3a = a + 3d
2a = 3d
a = (3/2)d
Similarly, for the 3rd term (n = 3),
a3 = a + (3 - 1)d
= a + 2d
Also, for the 7th term (n = 7),
a7 = a + (7 - 1)d
= a + 6d ...(3)
Now, using the value of a3 in equation (2), we get,
a7 = 2(a + 2d) + 1
= -2a + 4d + 1 ...(4)
Equating (3) and (4), we get,
a + 6d = 2a + 4d + 1
6d - 4d - 2a + a = +1
2d - a = +1
2d - (3/2)d = 1 (a = 3/2(d))
On further simplification, we get,
d = (1)(2)
d = 2
Now, to find a,
a = 3/2(d)
a = 3/2(2)
a = 3
Therefore, for the given A.P d = 2, a = 3
Ques.15. Find the second term and nth term of an A.P. whose 6th term is 12 and 8th term is 22.
Ans. In the given problem, we are given 6th and 8th term of an A.P.
We need to find the 2nd and nth term
Here, let us take the first term as a and the common difference as d
We are given,
a6 = 12
a8 = 22
Now, we will find a6 and a8 using the formula an = a + (n - 1)d
So,
a6 = a + (6 - 1)d
12 = a + 5d ...(1)
Also,
a8 = a + (8 - 1)d
22 = a + 7d ...(2)
So, to solve for a and d
On subtracting (1) from (2), we get
22 - 12 = (a +7d) - (a + 5d)
10 = a + 7d - a - 5d
10 = 2d
d = 10/2
d = 5 ...(3)
Substituting (3) in (1), we get
12 = a + 5(5)
a = 12 - 25
a = -13
Thus,
a = -13
d = 5
So, for the 2nd term (n = 2),
a2 = -13 + (2 - 1)5
= -13 + (1)5
= -13 + 5
= -8
For the nth term,
an = -13 + (n - 1)5
= -13 + 5n - 5
= -18 + 5n
Therefore, a2 = -8, an = 5n - 18
Ques.16. How many numbers of two digit are divisible by 3?
Ans. In this problem, we need to find out how many numbers of two digits are divisible by 3.
So, we know that the first two digit number that is divisible by 3 is 12 and the last two digit number divisible by 3 is 99. Also, all the terms which are divisible by 3 will form an A.P. with the common difference of 3.
So here,
First term (a) = 12
Last term (an) = 99
Common difference (d) = 3
So, let us take the number of terms as n
Now, as we know, an = a + (n - 1)d
So, for the last term,
99 = 12 + (n - 1)3
99 = 12 + 3n - 3
99 = 9 + 3n
99 - 9 = 3n
Further simplifying,
90 = 3n
n = 90/3
n = 30
Therefore, the number of two digit terms divisible by 3 is 30.
Ques.17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 32nd term.
Ans. In the given problem, we need to find the 32nd term of an A.P. which contains a total of 60 terms.
Here we are given the following,
First term (a) = 7
Last term (an) = 125
Number of terms (n) = 60
So, let us take the common difference as d
Now, as we know,
an = a + (n - 1)
So, for the last term,
125 = 7 + (60 - 1)d
125 = 7 + (59)d
125 - 7 = 59d
118 = 59d
Further simplifying,
d = 118/59
d = 2
So, for the 32nd term (n = 32)
a32 = 7 + (32 - 1)2
= 7 + (31)2
= 7 + 62
= 69
Therefore, the 32nd term of the given A.P. is 69.
Ques.18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.
Ans. In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 34.
We can write this as,
a4 + a8 = 24 ...(1)
a6 + a10 = 34 ...(2)
We need to find a and d
For the given A.P., let us take the first term as a and the common difference as d
As we know,
an = a + (n - 1)d
For 4th term (n = 4),
a4 = a + (4 - 1)d
= a + 3d
For 8th term (n = 8),
a8 = a + (8 - 1)d
= a + 7d
So, on substituting the above values in (1), we get,
(a + 3d) + (a + 7d) = 24
2a + 10d = 24 ...(3)
Also, for 6th term (n = 6),
a6 = a + (6 - 1)d
= a + 5d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
So, on substituting the above values in (2), we get,
(a + 5d)+(a + 9d) = 34
2a + 14d = 34 ...(4)
Next we simplify (3) and (4). On subtracting (3) from (4), we get,
(2a + 14d)-(2a + 10d) = 34 - 24
2a + 14d - 2a - 10d = 10
4d = 10
d = 10/4
d = 5/2
Further, using the value of d in equation (3), we get,
a + 10(5/2) = 24
2a + 5(5) = 24
2a + 25 = 24
2a = 24 - 25
On further simplifying, we get,
2a = -1
a = ((-1)/2)
Therefore, for the given A.P a = ((-1)/2) and d = 5/2.
Ques.19. The first term of an A.P. is 5 and its 100th term is −292. Find the 50th term of this A.P.
Ans. In the given problem, we are given 1st and 100th term of an A.P.
We need to find the 50th term
Here,
a = 5
a100 = -292
Now, we will find d using the formula an = a + (n - 1)d
So,
Also,
a100 = a + (100 - 1)d
-292 = a + 99d
So, to solve for d
Substituting a = 5, we get
-292 = 5 + 99d
-292 - 5 = 99d
((-297)/99) = d
d = -3
Thus,
a = 5, d = -3, n = 50
Substituting the above values in the formula, an = a + (n - 1)d
a50 = 5 + (50 - 1)(-3)
a50 = 5 - 147
a50 = -142
Therefore, a50 = -142.
Ques.20. Find a30 − a20 for the A.P.
(i) −9, −14, −19, −24, ...
(ii) a,a + d, a + 2d, a + 3d, ...
Ans. In this problem, we are given different A.P. and we need to find a30 - a20.
(i) A.P. −9, −14, −19, −24, ...
Here,
First term (a) = -9
Common difference of the A.P. (d) = -14 - (-9)
= -14 + 9
= -5
Now, as we know,
an = a + (n - 1)d
Here, we find a30 and a20
So, for 30th term,
a30 = a + (30 - 1)d
= -9 + (29)(-5)
= -9 - 145
= -154
Also, for 20th term,
a20 = a + (20 - 1)d
= -9 + (19)(-5)
= -9 - 95
= -104
So,
a30 - a20 = -154 - (-104)
= -154 + 104
= -50
Therefore, for the given A.P a30 - a20 = -50
(ii) A.P. a,a + d, a + 2d, a + 3d, ...
Here,
First term (a) = a
Common difference of the A.P. (d) = a + d - a = d
Now, as we know,
an = a + (n - 1)d
Here, we find a30 and a20.
So, for 30th term,
a30 = a + (30 - 1)d
= a + (29)d
Also, for 20th term,
a20 = a + (20 - 1)d
= a + (19)d
So,
a30 - a20 = (a + 29d)-(a + 19d)
= a + 29d - a - 19d
= 10d
Therefore, for the given A.P. a30 − a20 = 10d.
Ans. In the given problem, the 9th term of an A.P. is zero.
Here, let us take the first term of the A.P as a and the common difference as d
So, as we know,
an = a + (n - 1)d
We get,
a9 = a + (9 - 1)d
0 = a + 8 d
a = -8d ...(1)
Now, we need to prove that 29th term is double of 19th term. So, let us first find the two terms.
For 19th term (n = 19),
a19 = a + (19 - 1)d
= -8d + 18d (Using 1)
= 10d
For 29th term (n = 29),
a29 = a + (29 − 1)d (Using 1)
= −8d + 28d
= 20d
= 2 × 10d
=2 × a19
Therefore, for the given A.P. the 29th term is double of the 19th term.
Hence proved.
Ques.11. If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.
Ans. Here, let us take the first term of the A.P. as a and the common difference as d
We are given that 10 times the 10th term is equal to 15 times the 15th term. We need to show that 25th term is zero.
So, let us first find the two terms.
So, as we know,
an = a + (n - 1)d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
For 15th term (n = 15),
a15 = a + (15 - 1)d
= a + 14d
Now, we are given,
10(a + 9d) = 15(a + 14d)
Solving this, we get,
10a + 90d = 15a + 210d
90d - 210d = 15a - 10a
- 120d = 5a
-24d = a ...(i)
Next, we need to prove that the 25th term of the A.P. is zero. For that, let us find the 25th term using n = 25,
a25 = a + (25 - 1)d
= -24d + 24d (Using 1)
= 0
Thus, the 25th term of the given A.P. is zero.
Hence proved
Ques.12. In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Ans. Here, we are given that 24th term is twice the 10th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d
We have to prove that a72 = 2a34
So, let us first find the two terms.
As we know,
an = a + (n - 1)d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
For 24th term (n = 24),
a24 = a + (24 - 1)d
= a + 23d
Now, we are given that a24 = 2a10
So, we get,
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
23d - 18d = 2a - a
5d = a ...(1)
Further, we need to prove that the 72nd term is twice of 34th term. So let now find these two terms,
For 34th term (n = 34),
a34 = a + (34 - 1)d
= 5d + 33d (Using 1)
= 38d
For 72nd term (n = 72),
a72 = a + (72 - 1)d
= 5d + 71d (Using 1)
= 76d
= 2(38d)
Therefore, a72 = 2a34
Ques.13. The 26th, 11th and last term of an A.P. are 0, 3 and -(1/5), respectively. Find the common difference and the number of terms.
Ans. It is given that, a26 = 0, a11 = 3, an = −(1/5).
⇒ a + 25d = 0 .....(i)
⇒ a + 10d = 3 .....(ii)
Subtracting (ii) from (i).
15d = −3
⇒ d = −(1/5)
⇒ a = −25d = 5
Now, an = a +(n − 1)d
⇒ −(1/5) = 5 + (n − 1)(−(1/5))
⇒ −(1/5) - 5 = (n−1)(−(1/5))
⇒ n = 27
Ques.14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
Ans. In the given problem, let us take the first term as a and the common difference as d
Here, we are given that,
a4 = 3a ...(1)
a7 = 2a3 + 1 ...(2)
We need to find a and d
So, as we know,
an = a + (n - 1)d
For the 4th term (n = 4),
a4 = a + (4 - 1)d
3a = a + 3d
2a = 3d
a = (3/2)d
Similarly, for the 3rd term (n = 3),
a3 = a + (3 - 1)d
= a + 2d
Also, for the 7th term (n = 7),
a7 = a + (7 - 1)d
= a + 6d ...(3)
Now, using the value of a3 in equation (2), we get,
a7 = 2(a + 2d) + 1
= -2a + 4d + 1 ...(4)
Equating (3) and (4), we get,
a + 6d = 2a + 4d + 1
6d - 4d - 2a + a = +1
2d - a = +1
2d - (3/2)d = 1 (a = 3/2(d))
On further simplification, we get,
d = (1)(2)
d = 2
Now, to find a,
a = 3/2(d)
a = 3/2(2)
a = 3
Therefore, for the given A.P d = 2, a = 3
Ques.15. Find the second term and nth term of an A.P. whose 6th term is 12 and 8th term is 22.
Ans. In the given problem, we are given 6th and 8th term of an A.P.
We need to find the 2nd and nth term
Here, let us take the first term as a and the common difference as d
We are given,
a6 = 12
a8 = 22
Now, we will find a6 and a8 using the formula an = a + (n - 1)d
So,
a6 = a + (6 - 1)d
12 = a + 5d ...(1)
Also,
a8 = a + (8 - 1)d
22 = a + 7d ...(2)
So, to solve for a and d
On subtracting (1) from (2), we get
22 - 12 = (a +7d) - (a + 5d)
10 = a + 7d - a - 5d
10 = 2d
d = 10/2
d = 5 ...(3)
Substituting (3) in (1), we get
12 = a + 5(5)
a = 12 - 25
a = -13
Thus,
a = -13
d = 5
So, for the 2nd term (n = 2),
a2 = -13 + (2 - 1)5
= -13 + (1)5
= -13 + 5
= -8
For the nth term,
an = -13 + (n - 1)5
= -13 + 5n - 5
= -18 + 5n
Therefore, a2 = -8, an = 5n - 18
Ques.16. How many numbers of two digit are divisible by 3?
Ans. In this problem, we need to find out how many numbers of two digits are divisible by 3.
So, we know that the first two digit number that is divisible by 3 is 12 and the last two digit number divisible by 3 is 99. Also, all the terms which are divisible by 3 will form an A.P. with the common difference of 3.
So here,
First term (a) = 12
Last term (an) = 99
Common difference (d) = 3
So, let us take the number of terms as n
Now, as we know, an = a + (n - 1)d
So, for the last term,
99 = 12 + (n - 1)3
99 = 12 + 3n - 3
99 = 9 + 3n
99 - 9 = 3n
Further simplifying,
90 = 3n
n = 90/3
n = 30
Therefore, the number of two digit terms divisible by 3 is 30.
Ques.17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 32nd term.
Ans. In the given problem, we need to find the 32nd term of an A.P. which contains a total of 60 terms.
Here we are given the following,
First term (a) = 7
Last term (an) = 125
Number of terms (n) = 60
So, let us take the common difference as d
Now, as we know,
an = a + (n - 1)
So, for the last term,
125 = 7 + (60 - 1)d
125 = 7 + (59)d
125 - 7 = 59d
118 = 59d
Further simplifying,
d = 118/59
d = 2
So, for the 32nd term (n = 32)
a32 = 7 + (32 - 1)2
= 7 + (31)2
= 7 + 62
= 69
Therefore, the 32nd term of the given A.P. is 69.
Ques.18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.
Ans. In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 34.
We can write this as,
a4 + a8 = 24 ...(1)
a6 + a10 = 34 ...(2)
We need to find a and d
For the given A.P., let us take the first term as a and the common difference as d
As we know,
an = a + (n - 1)d
For 4th term (n = 4),
a4 = a + (4 - 1)d
= a + 3d
For 8th term (n = 8),
a8 = a + (8 - 1)d
= a + 7d
So, on substituting the above values in (1), we get,
(a + 3d) + (a + 7d) = 24
2a + 10d = 24 ...(3)
Also, for 6th term (n = 6),
a6 = a + (6 - 1)d
= a + 5d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
So, on substituting the above values in (2), we get,
(a + 5d)+(a + 9d) = 34
2a + 14d = 34 ...(4)
Next we simplify (3) and (4). On subtracting (3) from (4), we get,
(2a + 14d)-(2a + 10d) = 34 - 24
2a + 14d - 2a - 10d = 10
4d = 10
d = 10/4
d = 5/2
Further, using the value of d in equation (3), we get,
a + 10(5/2) = 24
2a + 5(5) = 24
2a + 25 = 24
2a = 24 - 25
On further simplifying, we get,
2a = -1
a = ((-1)/2)
Therefore, for the given A.P a = ((-1)/2) and d = 5/2.
Ques.19. The first term of an A.P. is 5 and its 100th term is −292. Find the 50th term of this A.P.
Ans. In the given problem, we are given 1st and 100th term of an A.P.
We need to find the 50th term
Here,
a = 5
a100 = -292
Now, we will find d using the formula an = a + (n - 1)d
So,
Also,
a100 = a + (100 - 1)d
-292 = a + 99d
So, to solve for d
Substituting a = 5, we get
-292 = 5 + 99d
-292 - 5 = 99d
((-297)/99) = d
d = -3
Thus,
a = 5, d = -3, n = 50
Substituting the above values in the formula, an = a + (n - 1)d
a50 = 5 + (50 - 1)(-3)
a50 = 5 - 147
a50 = -142
Therefore, a50 = -142.
Ques.20. Find a30 − a20 for the A.P.
(i) −9, −14, −19, −24, ...
(ii) a,a + d, a + 2d, a + 3d, ...
Ans. In this problem, we are given different A.P. and we need to find a30 - a20.
(i) A.P. −9, −14, −19, −24, ...
Here,
First term (a) = -9
Common difference of the A.P. (d) = -14 - (-9)
= -14 + 9
= -5
Now, as we know,
an = a + (n - 1)d
Here, we find a30 and a20
So, for 30th term,
a30 = a + (30 - 1)d
= -9 + (29)(-5)
= -9 - 145
= -154
Also, for 20th term,
a20 = a + (20 - 1)d
= -9 + (19)(-5)
= -9 - 95
= -104
So,
a30 - a20 = -154 - (-104)
= -154 + 104
= -50
Therefore, for the given A.P a30 - a20 = -50
(ii) A.P. a,a + d, a + 2d, a + 3d, ...
Here,
First term (a) = a
Common difference of the A.P. (d) = a + d - a = d
Now, as we know,
an = a + (n - 1)d
Here, we find a30 and a20.
So, for 30th term,
a30 = a + (30 - 1)d
= a + (29)d
Also, for 20th term,
a20 = a + (20 - 1)d
= a + (19)d
So,
a30 - a20 = (a + 29d)-(a + 19d)
= a + 29d - a - 19d
= 10d
Therefore, for the given A.P. a30 − a20 = 10d.
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FAQs on Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-4) - RD Sharma Solutions for Class 10 Mathematics
| 1. How do you solve quadratic equations by factoring? |  |
Ans. To solve quadratic equations by factoring, follow these steps:
1. Move all terms to one side of the equation to set it equal to zero.
2. Factor the quadratic expression.
3. Set each factor equal to zero and solve for the variable.
4. Check the solutions by substituting them back into the original equation.
| 2. What is the quadratic formula and how is it used to solve quadratic equations? | |
Ans. The quadratic formula is given by x = (-b ± √(b² - 4ac)) / 2a, where a, b, and c are coefficients of the quadratic equation ax² + bx + c = 0. To use the quadratic formula to solve quadratic equations, substitute the values of a, b, and c into the formula and simplify to find the roots of the equation.
| 3. Can all quadratic equations be solved using the quadratic formula? | |
Ans. Yes, all quadratic equations can be solved using the quadratic formula. The formula provides a method to find the roots of any quadratic equation, whether the equation is factorable or not.
| 4. How do you determine the nature of solutions of a quadratic equation using the discriminant? | |
Ans. The nature of solutions of a quadratic equation can be determined using the discriminant, which is given by the expression b² - 4ac. If the discriminant is greater than zero, the equation has two distinct real roots. If the discriminant is equal to zero, the equation has one real root. If the discriminant is less than zero, the equation has two complex roots.
| 5. Can quadratic equations have non-real solutions? | |
Ans. Yes, quadratic equations can have non-real solutions, which are complex numbers. When the discriminant of a quadratic equation is less than zero, the roots of the equation will be non-real or complex numbers.
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