Rd Chapter 9 Arithmetic Progressions Ex 9.4 Solutions
Question - 31 : - Which term of the arithmetic progression 8, 14, 20, 26, … will be 72 more than its 41st term ?
Answer - 31 : - In the given A.P. 8, 14, 20, 26, …
Question - 32 : - Find the term of the arithmetic progression 9, 12, 15, 18, … which is 39 more than its 36th term (C.B.S.E. 2006C)
Answer - 32 : -
In the given A.R 9,12, 15, 18, …
First term (a) = 9
and common difference (d) = 12 – 9 = 3
and an = a + (n – 1) d
Now a36 = a + (36 – 1) d = 9 + 35 x 3 = 9 + 105 = 114
Let the an be the required term
an = a + (n – 1) d
= 9 + (n – 1) x 3 = 9 + 3n – 3 = 6 + 3n
But their difference is 39
an – a36 = 39
=> 6 + 3n – 114 = 39
=> 114 – 6 + 39 = 3n
=> 3n = 147
=> n = 49
Required term is 49th
Question - 33 : - Find the 8th term from the end of the A.P. 7, 10, 13, …, 184. (C.B.S.E. 2005)
Answer - 33 : -
The given A.P. is 7,10, 13,…, 184
Here first term (a) = 7
and common difference (d) = 10 – 7 = 3
and last tenn (l) = 184
Let nth term from the last is an = l – (n – 1) d
a8= 184 – (8 – 1) x 3 = 184 – 7 x 3 = 184 – 21 = 163
Question - 34 : - Find the 10th term from the end of the A.P. 8, 10, 12, …, 126. (C.B.S.E. 2006)
Answer - 34 : -
The given A.P. is 8, 10, 12, …, 126
Here first term (a) = 8
Common difference (d) = 10 – 8 = 2
and last tenn (l) = 126
Now nth term from the last is an = l – (n – 1) d
a10 = 126 – (10 – 1) x 2 = 126 – 9 x 2 = 126 – 18 = 108
Question - 35 : - The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P. (C.B.S.E. 2009)
Answer - 35 : -
Question - 36 : - Which term of the A.P. 3, 15, 27, 39, …. will be 120 more than its 21st term ? (C.B.S.E. 2009)
Answer - 36 : -
A.P. is given : 3, 15, 27, 39, …….
Here first term (a) = 3
and c.d. (d) = 15 – 3 = 12
Let nth term be the required term
Now 21st term = a + (n – 1) d = 3 + 20 x 12 = 3 + 240 = 243
According to the given condition,
nth term – 21 st term = 120
=> a + (n – 1) d – 243 = 120
=> 3 + (n – 1) x 12 = 120 + 243 = 363
=> (n – 1) 12 = 363 – 3 = 360
=> n – 1 = 30
=> n = 30 + 1 = 31
31 st term is the required term
Question - 37 : - The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term.[CBSE 2012]
Answer - 37 : -
Question - 38 : - Find the number of ail three digit natural numbers which are divisible by 9. [CBSE 2013]
Answer - 38 : -
First 3-digit numberwhich is divisible by 9 = 108
and last 3-digit number = 999
d= 9
a + (n – 1) d = 999
=> 108 + (n – 1) x 9 = 999
=> (n – 1) d = 999 – 108
=> (n – 1) x 9 = 891
=> n – 1 = 99
=> n = 99 + 1 = 100
Number of terms = 100
Question - 39 : - The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P. [CBSE 2013]
Answer - 39 : -
Question - 40 : - The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P. [CBSE 2013]
Answer - 40 : -
Let a be the first term and d be the common difference and
Tn = a + (n – 1) d