Rd Chapter 9 Arithmetic Progressions Ex 9.4 Solutions
Question - 51 : - How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3? [NCERT Exemplar]
Answer - 51 : -
Here, the first numberis 11, which divided by 4 leave remainder 3 between 10 and 300.
Last term before 300 is 299, which divided by 4 leave remainder 3.
11, 15, 19, 23, …, 299
Here, first term (a) = 11,
common difference (d) = 15 – 11 = 4
nth term, an = a + (n – 1 ) d = l [last term]
=> 299 = 11 + (n – 1) 4
=> 299 – 11 = (n – 1) 4
=> 4(n – 1) = 288
=> (n – 1) = 72
n = 73
Answer - 52 : -
Given, A.P., -2, -4,-6, …, -100
Here, first term (a) = -2,
common difference (d) = -4 – (-2)
and the last term (l) = -100.
We know that, the nth term an of an A.P. from the end is an = l– (n – 1 )d,
where l is the last term and d is the common difference. 12th term from theend,
an = -100 – (12 – 1) (-2)
= -100 + (11) (2) = -100 + 22 = -78
Hence, the 12th term from the end is -78
Question - 53 : - For the A.P.: -3, -7, -11,…, can we find a30 – a20 without actually finding a30 and a20 ? Give reasons for your answer. [NCERT Exemplar]
Answer - 53 : -
True.
nth term of an A.P., an = a + (n – 1)d
a30 = a + (30 – 1 )d = a + 29d
and a20 = a + (20 – 1 )d = a + 19d …(i)
Now, a30 – a20 = (a + 29d) – (a + 19d) = 10d
and from given A.P.
common difference, d = -7 – (-3) = -7 + 3 = -4
a30 – a20 = 10(-4) = -40 [from Eq- (i)]
Question - 54 : - Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why? [NCERT
Answer - 54 : -
Let the same commondifference of two A.P.’s is d.
Given that, the first term of first A.P. and second A.P. are 2 and 7respectively,
then the A.P.’s are 2, 2 + d, 2 + 2d, 2 + 3d, … and 7, 7 + d, 7 + 2d, 7 + 3d, …
Now, 10th terms of first and second A.P.’s are 2 + 9d and 7 + 9d, respectively.
So, their difference is 7 + 9d – (2 + 9d) = 5
Also, 21st terms of first and second A.P.’s are 2 + 20d and 7 + 20d,respectively.
So, their difference is 7 + 20d – (2 + 9d) = 5
Also, if the an and bn are the nth terms offirst and second A.P.
Then bn – an = [7 + (n – 1 ) d] – [2 + (n – 1)d = 5
Hence, the difference between any two corresponding terms of such A.P.’s is thesame as the difference between their first terms.