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Rd Chapter 9 Arithmetic Progressions Ex 9.4 Solutions

Question - 1 : -
(i) 10th term of the A.P. 1, 4, 7, 10, ………
(ii) 18th term of the A.P. √2 , 3√2 , 5√2 , ……….
(iii) nth term of the A.P. 13, 8, 3, -2, ……..
(iv) 10th term of the A.P. -40, -15, 10, 35, ……..
(v) 8th term of the A.P. 117, 104, 91, 78, ………..
(vi) 11th term of the A.P. 10.0 , 10.5, 11.0, 11.5, ……….
(vii) 9th term of the A.P. 3/4 , 5/4 , 7/4 , 9/4 , ………

Answer - 1 : -


Question - 2 : -
(i) Which term of the A.P. 3, 8, 13, …… is 248 ?
(ii) Which term of the A.P. 84, 80, 76, ….. is 0 ?
(iii) Which term of the A.P. 4, 9, 14, ….. is 254 ?
(iv) Which term of the A.P. 21, 42, 63, 84, ….. is 420 ?
(v) Which term of the A.P. 121, 117, 113, ….. is its first negative term ?

Answer - 2 : -

(i) A.P. is 3, 8, 13, …, 248
Here first term (a) = 3
and common difference (d) = 8 – 3 = 5

Question - 3 : -
(i) Is 68 a term of the A.P. 7, 10, 13, …… ?
(ii) Is 302 a term of the A.P. 3, 8, 13, ….. ?
(ii) Is -150 a term of the A.P. 11, 8, 5, 2, …… ?

Answer - 3 : -


Question - 4 : -
How many terms are there in the A.P. ?
(i) 7, 10, 13, … 43
(ii) -1, – 5/6 , – 2/3 , – 1/2 , …….., 10/3
(iii) 7, 13, 19, …, 205
(iv) 18, 15(1/2) , 13, …, -47

Answer - 4 : -


Question - 5 : - The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.

Answer - 5 : -

The first term of anA.P. (a) = 5
and common difference (d) = 3
Last term = 80
Let the last term be nth
an = a + (n – 1) d
=> 80 = 5 + (n – 1) x 3
=> 80= 5 + 3n – 3
=> 3n = 80 – 5 + 3 = 78
=> n = 26
Number of terms = 26

Question - 6 : - The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

Answer - 6 : -

6th term of A.P. = 19
and 17th term = 41
Let a be the first term, and d be the common difference
We know that

Question - 7 : - If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.

Answer - 7 : -


Question - 8 : - If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.

Answer - 8 : - Let a, a + d, a + 2d, a + 3d, ……… be an A.P.
an = a + (n – 1) d
Now a10 = a + (10 – 1) d = a + 9d
and a15 = a + (15 – 1) d = a + 14d

Question - 9 : - The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Answer - 9 : -


Question - 10 : - In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Answer - 10 : -

Let a, a + d, a + 2d,a + 3d, …….. be an A.P.
an = a + (n – 1) d
10th (a10) = a + (10 – 1) d = a + 9d
and 24th term (a24) = a + (24 – 1) d = a + 23d
24th term = 2 x 10th term
a + 23d = 2 (a + 9d)
=> a + 23d = 2a + 18d
=> 2a – a = 23d – 18d
=> a = 5d ….(i)
Now 72nd term = a + (72 – 1)d = a + 71d
and 34th term = a + (34 – 1) d = a + 33d
Now a + 71d – 5d + 71d = 76d
and a + 33d = 5d+ 33d = 38d
76d = 2 x 38d
72th term = 2 (34th term) = twice of the 34th term
Hence proved.

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