Rd Chapter 9 Arithmetic Progressions Ex 9.4 Solutions
Question - 21 : - Write the expression an – ak for the A.P. a, a + d, a + 2d, ……
Hence, find the common difference of the A.P. for which
(i) 11th term is 5 and 13th term is 79.
(ii) a10 – a5 = 200
(iii) 20th term is 10 more than the 18th term.
Answer - 21 : - In the A.P. a, a + d, a + 2d, …..
Question - 22 : - Find n if the given value of x is the nth term of the given A.P.
Answer - 22 : - 
Question - 23 : - The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
Answer - 23 : -
Question - 24 : - Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.
Answer - 24 : - Let a, a + d, a + 2d, a + 3d, ………. be the A.P.
an = a + (n – 1) d
But a3 = 16
Question - 25 : - The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P. [CBSE 2004]
Answer - 25 : -
Let a, a + d, a + 2d,a + 3d, be the A.P.
Here a is the first term and d is the common difference
an = a + (n – 1) d
Now a7 = a + (7 – 1) d = a + 6d = 32 ….(i)
and a13 = a + (13 – 1) d = a + 12d = 62 ….(ii)
Subtracting (i) from (ii)
6d = 30
=> d = 5
a + 6 x 5 = 32
=> a + 30 = 32
=> a = 32 – 30 = 2
A.P. will be 2, 7, 12, 17, ………..
Question - 26 : - Which term of the A.P. 3, 10, 17, … will be 84 more than its 13th term ? [CBSE 2004]
Answer - 26 : -
Question - 27 : - Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ?
Answer - 27 : -
Question - 28 : - For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,… and 3, 10, 17, … are equal ? (C.B.S.E. 2008)
Answer - 28 : -
In the A.P. 63, 65,67, …
a = 63 and d = 65 – 63 = 2
an = a1 + (n – 1) d = 63 + (n – 1) x 2 = 63 +2n – 2 = 61 + 2n
and in the A.P. 3, 10, 17, …
a = 3 and d = 10 – 3 = 7
an = a + (n – 1) d = 3 + (n – 1) x 7 = 3 + 7n – 7 = 7n – 4
But both nth terms are equal
61 + 2n = 7n – 4
=> 61 + 4 = 7n – 2n
=> 65 = 5n
=> n = 13
n = 13
Question - 29 : - How many multiples of 4 lie between 10 and 250 ?
Answer - 29 : -
All the terms between10 and 250 are multiple of 4
First multiple (a) = 12
and last multiple (l) = 248
and d = 4
Let n be the number of multiples, then
an = a + (n – 1) d
=> 248 = 12 + (n – 1) x 4 = 12 + 4n – 4
=> 248 = 8 + 4n
=> 4n = 248 – 8 = 240
n = 60
Number of terms are = 60
Question - 30 : - How many three digit numbers are divisible by 7 ?
Answer - 30 : -
First three digit number is 100 and last three digit number is 999
In the sequence of the required three digit numbers which are divisible by 7, will be between
a = 105 and last number l = 994 and d = 7
Let n be the number of terms, then
an = a + (n – 1) d
994 = 105 + (n – 1) x 7
994 = 105 + 7n – 7
=> 7n = 994 – 105 + 7
=> 7n = 896
=> n = 128
Number of terms =128