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Rd Chapter 9 Arithmetic Progressions Ex 9.4 Solutions

Question - 41 : - The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term. [CBSE 2013]

Answer - 41 : -


Hence 72nd term = 4times of 15th term

Question - 42 : - Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. [CBSE 2014]

Answer - 42 : -

Numbers divisible byboth 2 and 5 are 110, 120, 130, ………. , 990
Here a = 110, x = 120 – 110 = 10
an = 990
As a + (n – 1) d = 990
110 + (n – 1) (10) = 990
(n – 1) (10) = 990 – 110 = 880
n – 1 = 88
n = 88 + 1 = 89

Question - 43 : - If the seventh term of an AP is 1/9 and its ninth term is 1/7 , find its (63) rd term. [CBSE 2014]

Answer - 43 : -


Question - 44 : - The sum of 5th and 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP. [CBSE 2014]

Answer - 44 : -


Question - 45 : - Find where 0 (zero) is a term of the AP 40, 37, 34, 31, …… [CBSE 2014]

Answer - 45 : -

AP 40, 37, 34, 31, …..
Here a = 40, d = -3
Let Tn = 0
Tn = a + (n – 1) d
=> 0 = 40 + (n – 1) (-3)
=> 0 = 40 – 3n + 3
=> 3n = 43
=> n = 433 which is infraction
There is no term which is 0

Question - 46 : - Find the middle term of the A.P. 213, 205, 197, …, 37. [CBSE2015]

Answer - 46 : -


Question - 47 : - If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P. [BTE2015]

Answer - 47 : -

We know that,
Tn = a + (n – 1 )d
T5 = a + 4d => a + 4d = 31 ……(i)
and T25 = a + 24d
=>a + 24d = 140 + T5
=> a + 24d = 140 + 31 = 171 …..(ii)
Subtracting (i) from (ii),
20d= 140
and a + 4d = 31
=> a + 4 x 7 = 31
=> a + 28 = 31
=> a = 31 – 28 = 3
a = 3 and d = 7
AP will be 3, 10, 17, 24, 31, ……..

Question - 48 : - Find the sum of two middle terms of the

Answer - 48 : -


Question - 49 : - If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.

Answer - 49 : -


Question - 50 : - If an A.P. consists of n terms with first term a and nth term l show that the sum of the mth term from the beginning and the mth term from the end is (a + l).

Answer - 50 : -

In an A.P.
Number of terms = n
First term = a
and nth term = l
mth term (am) = a + (m – 1) d
and mth term from the end = l – (m – 1)d
Their sum = a + (m – 1) d + l – (m – 1) d = a + l
Hence proved.

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