Q1: Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Ans: Assume that A_{1}, A_{2}, A_{3}, A_{4}, and A_{5} are the five numbers between 8 and 26, such that the sequence of an A.P becomes 8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 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_{1 }= α + d = 8 + 3 =11
A_{2 }= α + 2d = 8 + 2(3) = 8 + 6 = 14
A_{3 }= α + 3d = 8 + 3(3) = 8 + 9 = 17
A_{4 }= α + 4d = 8 + 4(3) = 8 + 12 = 20
A_{5 }= 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, α_{m+n} = α +(m+n -1)d
α_{m-n} = α +(m-n -1)d
α_{m} = a +(m-1)d
Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:
α_{m+n }+ α_{m-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 α_{m} [since α_{m} = 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 α_{1}, α_{2} and d_{1}, d_{2} 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)[2α_{1}+ (n-1)d_{1}]]/ [(n/2)[2α_{2 }+ (n-1)d_{2}]]= (5n+4)/(9n+6)
Cancel out (n/2) both numerator and denominator on L.H.S
⇒ [2α_{1 }+ (n-1)d_{1}]/[2a_{2 }+ (n-1)d_{2}] = (5n+4)/(9n+6) …(1)
Now substitute n= 35 in equation (1), {Since (n-1)/2 = 17}
Then equation (1) becomes
⇒ [2α_{1 }+ 34d_{1}]/[2α_{2}+ 34d_{2}]= (5(35)+4)/(9(35+6)
⇒ [α_{1}+ 17d_{1}]/[α_{2}+ 17d_{2}]= 179/321 …(2)
Now, we can say that.
18th term of first AP/ 18th term of second AP = [α_{1}+ 17d_{1}]/[a_{2}+ 17d_{2}]….(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 q^{2} = ps
Ans: Given that:
5th term = P
8th term = q
11th term = s
To prove that: q^{2} = ps
By using the above information, we can write the equation as:
α_{5} = αr^{5-1} = αr^{4} = p ….(1)
α_{8} = αr^{8-1} = αr^{7} = q ….(2)
α_{11 }= αr^{11-1} = αr^{10} =s …(3)
Divide the equation (2) by (1), we get
r^{3} = q/p …(4)
Divide the equation (3) by (2), we get
r^{3} = s/q …(5)
Now, equate the equation (4) and (5), we get
q/p = s/q
It becomes, q^{2} = 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.