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, тАж, 299
Here, first term (a) = 11,
common difference (d) = 15 тАУ 11 = 4
nth term, an┬а= a + (n тАУ 1 ) d = l [last term]
=> 299 = 11 + (n тАУ 1) 4
=> 299 тАУ 11 = (n тАУ 1) 4
=> 4(n тАУ 1) = 288
=> (n тАУ 1) = 72
n = 73
Answer - 52 : -
Given, A.P., -2, -4,-6, тАж, -100
Here, first term (a) = -2,
common difference (d) = -4 тАУ (-2)
and the last term (l) = -100.
We know that, the nth term an of an A.P. from the end is an┬а= lтАУ (n тАУ 1 )d,
where l is the last term and d is the common difference. 12th term from theend,
an┬а= -100 тАУ (12 тАУ 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,тАж, can we find a30 тАУ a20 without actually finding a30 and a20 ? Give reasons for your answer. [NCERT Exemplar]
Answer - 53 : -
True.
nth term of an A.P., an┬а= a + (n тАУ 1)d
a30┬а= a + (30 тАУ 1 )d = a + 29d
and a20┬а= a + (20 тАУ 1 )d = a + 19d тАж(i)
Now, a30┬атАУ a20┬а= (a + 29d) тАУ (a + 19d) = 10d
and from given A.P.
common difference, d = -7 тАУ (-3) = -7 + 3 = -4
a30┬атАУ a20┬а= 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┬а
Answer - 54 : -
Let the same commondifference of two A.P.тАЩs is d.
Given that, the first term of first A.P. and second A.P. are 2 and 7respectively,
then the A.P.тАЩs are 2, 2 + d, 2 + 2d, 2 + 3d, тАж and 7, 7 + d, 7 + 2d, 7 + 3d, тАж
Now, 10th terms of first and second A.P.тАЩs are 2 + 9d and 7 + 9d, respectively.
So, their difference is 7 + 9d тАУ (2 + 9d) = 5
Also, 21st terms of first and second A.P.тАЩs are 2 + 20d and 7 + 20d,respectively.
So, their difference is 7 + 20d тАУ (2 + 9d) = 5
Also, if the an┬аand bn┬аare the nth terms offirst and second A.P.
Then bn┬атАУ an┬а= [7 + (n тАУ 1 ) d] тАУ [2 + (n тАУ 1)d = 5
Hence, the difference between any two corresponding terms of such A.P.тАЩs is thesame as the difference between their first terms.