RD Chapter 19 Arithmetic Progressions Ex 19.2 Solutions
Question - 1 : - Find:
(i) 10th term of the A.P. 1, 4, 7, 10, …..
(ii) 18th term of the A.P. √2, 3√2, 5√2, …
(iii) nth term of the A.P 13, 8, 3, -2, ….
Answer - 1 : -
(i) 10th termof the A.P. 1, 4, 7, 10, …..
Arithmetic Progression(AP) whose common difference is = an – an-1 wheren > 0
Let us consider, a = a1 =1, a2 = 4 …
So, Common difference,d = a2 – a1 = 4 – 1 = 3
To find the 10th termof A.P, firstly find an
By using the formula,
an = a+ (n-1) d
= 1 + (n-1) 3
= 1 + 3n – 3
= 3n – 2
When n = 10:
a10 =3(10) – 2
= 30 – 2
= 28
Hence, 10th termis 28.
(ii) 18th termof the A.P. √2, 3√2, 5√2, …
Arithmetic Progression(AP) whose common difference is = an – an-1 wheren > 0
Let us consider, a = a1 =√2, a2 = 3√2 …
So, Common difference,d = a2 – a1 = 3√2 – √2 = 2√2
To find the 18th termof A.P, firstly find an
By using the formula,
an = a+ (n-1) d
= √2 + (n – 1) 2√2
= √2 + 2√2n – 2√2
= 2√2n – √2
When n = 18:
a18 =2√2(18) – √2
= 36√2 – √2
= 35√2
Hence, 10th termis 35√2
(iii) nth term of the A.P13, 8, 3, -2, ….
Arithmetic Progression(AP) whose common difference is = an – an-1 wheren > 0
Let us consider, a = a1 =13, a2 = 8 …
So, Common difference,d = a2 – a1 = 8 – 13 = -5
To find the nth termof A.P, firstly find an
By using the formula,
an = a+ (n-1) d
= 13 + (n-1) (-5)
= 13 – 5n + 5
= 18 – 5n
Hence, nth termis 18 – 5n
Question - 2 : - In an A.P., show that am+n + am–n = 2am.
Answer - 2 : -
We know the first termis ‘a’ and the common difference of an A.P is d.
Given:
am+n +am–n = 2am
By using the formula,
an = a+ (n – 1)d
Now, let us take LHS:am+n + am-n
am+n +am-n = a + (m + n – 1)d + a + (m – n – 1)d
= a + md + nd – d + a+ md – nd – d
= 2a + 2md – 2d
= 2(a + md – d)
= 2[a + d(m – 1)] {∵ an = a + (n – 1)d}
am+n +am-n = 2am
Hence Proved.
Question - 3 : - (i) Which term of the A.P. 3, 8, 13,… is 248 ?
(ii) Which term of the A.P. 84, 80, 76,… is 0 ?
(iii) Which term of the A.P. 4, 9, 14,… is 254 ?
Answer - 3 : -
(i) Which term of the A.P.3, 8, 13,… is 248 ?
Given A.P is 3, 8,13,…
Here, a1 =a = 3, a2 = 8
Common difference, d =a2 – a1 = 8 – 3 = 5
We know, an =a + (n – 1)d
an = 3+ (n – 1)5
= 3 + 5n – 5
= 5n – 2
Now, to find whichterm of A.P is 248
Put an =248
∴ 5n – 2 = 248
= 248 + 2
= 250
= 250/5
= 50
Hence, 50th termof given A.P is 248.
(ii) Which term of the A.P.84, 80, 76,… is 0 ?
Given A.P is 84, 80,76,…
Here, a1 =a = 84, a2 = 88
Common difference, d =a2 – a1 = 80 – 84 = -4
We know, an =a + (n – 1)d
an =84 + (n – 1)-4
= 84 – 4n + 4
= 88 – 4n
Now, to find whichterm of A.P is 0
Put an =0
88 – 4n = 0
-4n = -88
n = 88/4
= 22
Hence, 22nd termof given A.P is 0.
(iii) Which term of the A.P.4, 9, 14,… is 254 ?
Given A.P is 4, 9,14,…
Here, a1 =a = 4, a2 = 9
Common difference, d =a2 – a1 = 9 – 4 = 5
We know, an =a + (n – 1)d
an = 4+ (n – 1)5
= 4 + 5n – 5
= 5n – 1
Now, to find whichterm of A.P is 254
Put an =254
5n – 1 = 254
5n = 254 + 1
5n = 255
n = 255/5
= 51
Hence, 51st termof given A.P is 254.
Question - 4 : - (i) Is 68 a term of the A.P. 7, 10, 13,…?
(ii) Is 302 a term of the A.P. 3, 8, 13,…?
Answer - 4 : -
(i) Is 68 a term of theA.P. 7, 10, 13,…?
Given A.P is 7, 10,13,…
Here, a1 =a = 7, a2 = 10
Common difference, d =a2 – a1 = 10 – 7 = 3
We know, an =a + (n – 1)d [where, a is first term or a1 and d is commondifference and n is any natural number]
an = 7+ (n – 1)3
= 7 + 3n – 3
= 3n + 4
Now, to find whether68 is a term of this A.P. or not
Put an =68
3n + 4 = 68
3n = 68 – 4
3n = 64
n = 64/3
64/3 is not anatural number
Hence, 68 is not aterm of given A.P.
(ii) Is 302 a term of theA.P. 3, 8, 13,…?
Given A.P is 3, 8,13,…
Here, a1 =a = 3, a2 = 8
Common difference, d =a2 – a1 = 8 – 3 = 5
We know, an =a + (n – 1)d
an = 3+ (n – 1)5
= 3 + 5n – 5
= 5n – 2
To find whether 302 isa term of this A.P. or not
Put an =302
5n – 2 = 302
5n = 302 + 2
5n = 304
n = 304/5
304/5 is not anatural number
Hence, 304 is not aterm of given A.P.
Question - 5 : - (i) Which term of the sequence 24, 23 ¼, 22 ½, 21 ¾ is the firstnegative term?
(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, … is (a)purely real (b) purely imaginary ?
Answer - 5 : -
(i) Given:
AP: 24, 23 ¼, 22 ½, 21¾, … = 24, 93/4, 45/2, 87/4, …
Here, a1 =a = 24, a2 = 93/4
Common difference, d =a2 – a1 = 93/4 – 24
= (93 – 96)/4
= – 3/4
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
We know, an =a + (n – 1) d
an =24 + (n – 1) (-3/4)
= 24 – 3/4n + ¾
= (96+3)/4 – 3/4n
= 99/4 – 3/4n
Now we need to find,first negative term.
Put an <0
an =99/4 – 3/4n < 0
99/4 < 3/4n
3n > 99
n > 99/3
n > 33
Hence, 34th termis the first negative term of given AP.
(ii) Given:
AP: 12 + 8i, 11 + 6i,10 + 4i, …
Here, a1 =a = 12 + 8i, a2 = 11 + 6i
Common difference, d =a2 – a1
= 11 + 6i – (12 + 8i)
= 11 – 12 + 6i – 8i
= -1 – 2i
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
an =12 + 8i + (n – 1) -1 – 2i
= 12 + 8i – n – 2ni +1 + 2i
= 13 + 10i – n – 2ni
= (13 – n) + (10 – 2n)i
To find purely realterm of this A.P., imaginary part have to be zero
10 – 2n = 0
2n = 10
n = 10/2
= 5
Hence, 5th termis purely real.
To find purelyimaginary term of this A.P., real part have to be zero
∴ 13 – n = 0
n = 13
Hence, 13th termis purely imaginary.
Question - 6 : - (i) How many terms are in A.P. 7, 10, 13,…43?
(ii) How many terms are there in the A.P. -1, -5/6, -2/3, -1/2, …, 10/3 ?
Answer - 6 : -
(i) Given:
AP: 7, 10, 13,…
Here, a1 =a = 7, a2 = 10
Common difference, d =a2 – a1 = 10 – 7 = 3
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
an = 7+ (n – 1)3
= 7 + 3n – 3
= 3n + 4
To find total terms ofthe A.P., put an = 43 as 43 is last term of A.P.
3n + 4 = 43
3n = 43 – 4
3n = 39
n = 39/3
= 13
Hence, total 13 termsexists in the given A.P.
(ii) Given:
AP: -1, -5/6, -2/3,-1/2, …
Here, a1 =a = -1, a2 = -5/6
Common difference, d =a2 – a1
= -5/6 – (-1)
= -5/6 + 1
= (-5+6)/6
= 1/6
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
an =-1 + (n – 1) 1/6
= -1 + 1/6n – 1/6
= (-6-1)/6 + 1/6n
= -7/6 + 1/6n
To find total terms ofthe AP,
Put an =10/3 [Since, 10/3 is the last term of AP]
an =-7/6 + 1/6n = 10/3
1/6n = 10/3 + 7/6
1/6n = (20+7)/6
1/6n = 27/6
n = 27
Hence, total 27 termsexists in the given A.P.
Question - 7 : - The first term of an A.P. is 5, the common difference is 3, and the lastterm is 80; find the number of terms.
Answer - 7 : -
Given:
First term, a = 5;last term, l = an = 80
Common difference, d =3
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
an = 5+ (n – 1)3
= 5 + 3n – 3
= 3n + 2
To find total terms ofthe A.P., put an = 80 as 80 is last term of A.P.
3n + 2 = 80
3n = 80 – 2
3n = 78
n = 78/3
= 26
Hence, total 26 termsexists in the given A.P.
Question - 8 : - The 6th and 17th terms of an A.P. are 19and 41 respectively. Find the 40th term.
Answer - 8 : -
Given:
6th termof an A.P is 19 and 17th terms of an A.P. is 41
So, a6 =19 and a17 = 41
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
When n = 6:
a6 = a+ (6 – 1) d
= a + 5d
Similarly, When n= 17:
a17 =a + (17 – 1)d
= a + 16d
According to question:
a6 =19 and a17 = 41
a + 5d = 19 ……………… (i)
And a + 16d = 41…………..(ii)
Let us subtractequation (i) from (ii) we get,
a + 16d – (a + 5d) =41 – 19
a + 16d – a – 5d = 22
11d = 22
d = 22/11
= 2
put the value of d inequation (i):
a + 5(2) = 19
a + 10 = 19
a = 19 – 10
= 9
As, an =a + (n – 1)d
a40 =a + (40 – 1)d
= a + 39d
Now put the value of a= 9 and d = 2 in a40 we get,
a40 =9 + 39(2)
= 9 + 78
= 87
Hence, 40th termof the given A.P. is 87.
Question - 9 : - If 9th term of an A.P. is Zero, prove that its 29th termis double the 19th term.
Answer - 9 : -
Given:
9th termof an A.P is 0
So, a9 =0
We need to prove: a29 =2a19
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
When n = 9:
a9 = a+ (9 – 1)d
= a + 8d
According to question:
a9 = 0
a + 8d = 0
a = -8d
When n = 19:
a19 =a + (19 – 1)d
= a + 18d
= -8d + 18d
= 10d
When n = 29:
a29 =a + (29 – 1)d
= a + 28d
= -8d + 28d[Since, a = -8d]
= 20d
= 2×10d
a29 =2a19 [Since, a19 = 10d]
Hence Proved.
Question - 10 : - If 10 times the 10th term of an A.P. is equal to 15 timesthe 15th term, show that the 25th term of theA.P. is Zero.
Answer - 10 : -
Given:
10 times the 10th termof an A.P. is equal to 15 times the 15th term
So, 10a10 =15a15
We need to prove: a25 =0
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
When n = 10:
a10 =a + (10 – 1)d
= a + 9d
When n = 15:
a15 =a + (15 – 1)d
= a + 14d
When n = 25:
a25 =a + (25 – 1)d
= a + 24d ………(i)
According to question:
10a10 =15a15
10(a + 9d) = 15(a +14d)
10a + 90d = 15a + 210d
10a – 15a + 90d – 210d= 0
-5a – 120d = 0
-5(a + 24d) = 0
a + 24d = 0
a25 =0 [From (i)]
Hence Proved.