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Important Questions: Sequences & Series | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Ans: 
Assume that A1, A2, A3, A4, and A5 are the five numbers between 8 and 26, such that the sequence of an A.P becomes 8, A1, A2, A3, A4, A5, 26
Here, a = 8, l = 26, n = 5
Therefore, 26= 8+(7-1)d
Hence it becomes,
26 = 8 + 6d
6d = 26-8 = 18
6d = 18
d = 3
A= α + d = 8 + 3 =11
A= α + 2d = 8 + 2(3) = 8 + 6 = 14
A= α + 3d = 8 + 3(3) = 8 + 9 = 17
A= α + 4d = 8 + 4(3) = 8 + 12 = 20
A= α +5d = 8 + 5(3) = 8 + 15 = 23
Hence, the required five numbers between the number 8 and 26 are 11, 14, 17, 20, 23.

Q2: Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Ans: 
Let a and d be the first term and the common difference of the A.P. respectively. It is known
that the kth term of an A.P. is given by
αk = α +(k -1)d
Therefore, αm+n = α +(m+n -1)d
αm-n = α +(m-n -1)d
αm = a +(m-1)d
Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:
αm+n + αm-n = α +(m+n -1)d + α +(m-n -1)d
= 2α +(m + n -1+ m – n -1)d
=2α + (2m-2)d
=2α + 2(m-1)d
= 2 [α + (m-1)d]
= 2 αm [since αm = α +(m-1)d]
Therefore, the sum of (m + n) and (m – n)th terms of an A.P. is equal to twice the mth term.

Q3: The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms.
Ans: 
Let α1, α2 and d1, d2 be the first term and the common difference of the first and second arithmetic progression respectively.
Then,
(Sum of n terms of the first A.P)/(Sum of n terms of the second A.P) = (5n+4)/(9n+6)
⇒ [ (n/2)[2α1+ (n-1)d1]]/ [(n/2)[2α+ (n-1)d2]]= (5n+4)/(9n+6)
Cancel out (n/2) both numerator and denominator on L.H.S
⇒ [2α+ (n-1)d1]/[2a+ (n-1)d2] = (5n+4)/(9n+6) …(1)
Now substitute n= 35 in equation (1), {Since (n-1)/2 = 17}
Then equation (1) becomes
⇒ [2α+ 34d1]/[2α2+ 34d2]= (5(35)+4)/(9(35+6)
⇒ [α1+ 17d1]/[α2+ 17d2]= 179/321 …(2)
Now, we can say that.
18th term of first AP/ 18th term of second AP = [α1+ 17d1]/[a2+ 17d2]….(3)
Now, from (2) and (3), we can say that,
18th term of first AP/ 18th term of second AP = 179/321
Hence, the ratio of the 18th terms of both the AP’s is 179:321.

Q4: The 5th, 8th, and 11th terms of a GP are p, q and s respectively. Prove that q2 = ps
Ans:
Given that:
5th term = P
8th term = q
11th term = s
To prove that: q2 = ps
By using the above information, we can write the equation as:
α5 = αr5-1 = αr4 = p ….(1)
α8 = αr8-1 = αr7 = q ….(2)
α11 = αr11-1 = αr10 =s …(3)
Divide the equation (2) by (1), we get
r3 = q/p …(4)
Divide the equation (3) by (2), we get
r3 = s/q …(5)
Now, equate the equation (4) and (5), we get
q/p = s/q
It becomes, q2 = ps
Hence proved.

Q5: Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Ans: 
The integers from 1 to 100, which are divisible by 2, are 2, 4, 6 ….. 100.
This forms an A.P. with both the first term and common difference equal to 2.
⇒ 100=2+(n-1)2
⇒ n= 50
Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:
2+4+6+…+100 = (50/2)[2(2)+(50-1)(2)]
= (50/2)(4+98)
= 25(102)
= 2550
The integers from 1 to 100, which are divisible by 5, 10…. 100
This forms an A.P. with both the first term and common difference equal to 5.
Therefore, 100= 5+(n-1)5
⇒5n = 100
⇒ n= 100/5
⇒ n= 20
Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:
5+10+15+…+100= (20/2)[2(5)+(20-1)(5)]
= (20/2)(10+95)
= 10(105)
= 1050
Hence, the integers from 1 to 100, which are divisible by both 2 and 5 are 10, 20, ….. 100.
This also forms an A.P. with both the first term and common difference equal to 10.
Therefore, 100= 10+(n-1)10
⇒10n = 100
⇒ n= 100/10
⇒ n= 10
10+20+…+100= (10/2)[2(10)+(10-1)(10)]
= (10/2)(20+90)
= 5(110)
= 550
Therefore, the required sum is:
= 2550+ 1050 – 550
= 3050
Hence, the sum of the integers from 1 to 100, which are divisible by 2 or 5 is 3050.

The document Important Questions: Sequences & Series | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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FAQs on Important Questions: Sequences & Series - Mathematics (Maths) Class 11 - Commerce

1. What are sequences and series?
Ans. Sequences are ordered lists of numbers, while series are the sums of these numbers. In a sequence, each element is obtained by adding a common difference to the previous element, while in a series, the elements of a sequence are added together.
2. What is the difference between an arithmetic and geometric sequence?
Ans. In an arithmetic sequence, each term is obtained by adding a constant difference to the previous term. In a geometric sequence, each term is obtained by multiplying the previous term by a common ratio.
3. How do you find the nth term of an arithmetic sequence?
Ans. To find the nth term of an arithmetic sequence, you can use the formula: nth term = first term + (n - 1) * common difference. Here, the first term is the initial value of the sequence, n is the position of the term, and the common difference is the constant value added to each term.
4. What is the formula for the sum of an arithmetic series?
Ans. The formula for the sum of an arithmetic series is: sum = (n/2) * (2a + (n - 1) * d), where n is the number of terms, a is the first term, and d is the common difference.
5. How do you find the sum of an infinite geometric series?
Ans. The sum of an infinite geometric series can be found using the formula: sum = a / (1 - r), where a is the first term and r is the common ratio. However, this formula only works if the absolute value of r is less than 1, ensuring convergence.
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