1. Find the sum of all natural numbers lying between 100 and 1000, multiples of 5.
Solution:
The natural numbers between 100 and 1000, multiples of 5, are 105, 110, … 995.
It forms a sequence in A.P.
Where, a = 10, d = 5
Now,
a + (n -1)d = 995
105 + (n – 1)(5) = 995
5n = 995 – 105 + 5 = 895
n = 895/5=179
We know,
S_{n} = n/2 [2a + (n-1)d]
Therefore, the sum of all natural numbers lying between 100 and 1000, multiples of 5, is 98450.
2. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
Solution:
Given A.P.=25, 22, 19, …
Here, a = 25, d = 22 – 25 = -3
S_{n}= 116
The number of terms be n
So, we have
S_{n} = n/2 [2a + (n-1)d] = 116
116 = n/2 [2(25) + (n-1)(-3)]
232 = 53n – 3n^{2}
3n^{2} – 53n + 232 = 0
3n(n – 8) – 29(n – 8) = 0
(3n – 29) (n – 8) = 0
Hence,
n = 29/3 or n = 8
As n can only be an integral value, n = 8
Thus, 8th term is the last term of the A.P.
a_{8} = 25 + (8 – 1)(-3)
= 25 – 21
= 4
3. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Solution:
Let’s assume A1, A2, A3, A4, and A5 to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.
Here we have,
a = 8, b = 26, n = 7
So,
26 = 8 + (7 – 1) d
6d = 26 – 8 = 18
d = 3
Now,
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23
Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.
4. A man starts repaying a loan as the first installment of Rs. 100. If he increases the installment by Rs 5 every month, what will he pay in the 30th installment?
Solution:
Given,
The first installment of the loan is Rs 100.
The second installment of the loan is Rs 105 as the installment increases by Rs 5 every month.
Thus, the amount that the man repays every month forms an A.P.
And the A.P. is 100, 105, 110, …
Where, the first term, a = 100
Common difference, d = 5
So, the 30th term in this A.P. will be
A_{30} = a + (30 – 1)d
= 100 + (29) (5)
= 100 + 145
= 245
Therefore, the amount to be paid in the 30th installment will be Rs 245.
5. If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.
Solution:
Given, a, b, c, d are in G.P.
So, we have
bc = ad … (1)
b^{2} = ac … (2)
c^{2} = bd … (3)
Taking the R.H.S. we have
R.H.S.
= (ab + bc + cd)^{2}
= (ab + ad + cd)^{2 } [Using (1)]
= [ab + d (a + c)]^{2}
= a^{2}b^{2} + 2abd (a + c) + d^{2} (a + c)^{2}
= a^{2}b^{2} +2a^{2}bd + 2acbd + d^{2}(a^{2} + 2ac + c^{2})
= a^{2}b^{2} + 2a^{2}c^{2} + 2b^{2}c^{2} + d^{2}a^{2} + 2d^{2}b^{2} + d^{2}c^{2} [Using (1) and (2)]
= a^{2}b^{2}+ a^{2}c^{2} + a^{2}c^{2} + b^{2}c^{2} + b^{2}c^{2} + d^{2}a^{2} + d^{2}b^{2} + d^{2}b^{2} + d^{2}c^{2}
= a^{2}b^{2}+ a^{2}c^{2} + a^{2}d^{2} + b^{2} × b^{2} + b^{2}c^{2} + b^{2}d^{2} + c^{2}b^{2} + c^{2} × c^{2} + c^{2}d^{2}
[Using (2) and (3) and rearranging terms]
= a^{2}(b^{2} + c^{2} + d^{2}) + b^{2} (b^{2} + c^{2} + d^{2}) + c^{2} (b^{2}+ c^{2} + d^{2})
= (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})
= L.H.S.
Thus, L.H.S. = R.H.S.
Therefore,
(a^{2} + b^{2} + c^{2})(b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2}