Sunday, 4 November 2012

CBSE I NCERT 10th Arithmetic Progression Problems with solution

Solved Arithmetic Progression Problems By Jsunil
(1) Determine k so that k+2 , 4k-6 and 3k-2 are the three consecutive terms of an AP.
a1= k+2
a2= 4k-6
a3= 3k-2
a2-a1= a3- a2
4k-6-(k+2) = 3k-2-(4k-6)
4k -6-k-2 = 3k-2-4k+6
3k-8 = -k+4
3k+k = 4+8
4k = 12
k = 3
(2) if 7 times the 7th term of an AP is equal to 11 times the 11th term , show that the 18th term is zero.
Given: 7 times the 7th term of an AP is equal to 11 times the 11th term
7(a+6d) =11(a+10d)
7a+42d=11a+110d
42d-110d=11a-7a
68d = 4a
a =-17d
Now, the 18th term = a+17d=-17d-17d=0
(3) If the nth term of an A.P is 7n-5. Find 100th term
Given that the n th term of the A.P. is 7n-5
So 100 th term will be 7 (100) -5 =695

(4) if m times the mth term of an AP is equal to n times the nth term . Show that (m+n)th term of the AP is zero
We know :- an = a +(n-1)d
(m+n) = a + (m+n-1)d (just put m+n in place of n ) ------------------------------(1)
Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.
Then, m th term = a + (m – 1) d and n th term = a + (n – 1) d
By the given condition,
m x am = n x an
m [a + (m – 1) d] = n [a + (n – 1) d]
 ma + m (m – 1) d = na + n (n – 1) d
=> ma + (m2 -m)d - na - (n2 -n)d = 0 ( taking the Left Hand Side to the other side )
=> ma -na + (m2 - m)d -( n2-n)d = 0 (re-ordering the terms)
=> a (m-n) + d (m2-n2-m+n) = 0 (taking 'a ' and 'd ' common)
=> a (m-n) + d {(m+n)(m-n)-(m-n)} = 0 (a2-b2 identity)
Now divide both sides by (m-n)
=> a (1) + d {(m+n)(1)-(1)} = 0
=>a + d (m+n-1) = 0 ---------------(ii)
From equation number 1 and 2 ,
(m+n) = a + (m+n-1)d
And we have shown ,
a + d (m+n-1) = 0
So, a (m+n) = 0 
(5). Prove that the nth term of an AP cannot be n2 + 1. Justify your answer.
Common difference of an A.P. must always be a constant.
 d cannot be n – 1. Here, d varies when n takes different values.
For n = 1, d = 1 – 1 = 0
For n = 2, d = 2 – 1 = 1
For n = 3, d = 3 – 1 = 2
 d is not constant.
Thus, d cannot be taken as n – 1.
an is the n th term of an A.P. if an – an –1 = constant
Given, an n 2 + 1
an – an –1 = (n 2 + 1) – [(n – 1)2 + 1]
= (n 2 + 1) – (n 2 – 2n + 2)
= 2n – 1        
 an – an –1 ≠ constant
Thus, an n 2 + 1 cannot be the n th term of A.P.
(6) Find the sum of the first k terms of a series whose n th term is 2an+b 
The n th term of the AP is given by 2an+b
a1=2a+b
a2 =4a+b
a3=6a+b
Common difference = d=( 4a + b ) - ( 2a + b ) = 2a
Therefore, sum of first k terms =k/2[(2a+(k-1)d]
 = k/2[(2(2a+b)+(k-1)2a]= k/2   x 2 (2a+b+k-a)=k(a+b+ak)
(7) Which term of the AP,  3,10,17 will be 84 more than its 13th term?
Let the nth term be 84 more than the 13th term.
Now a/q,
a=3, d=10-3=7
So, 13th term= a+12d =3+12x7=87
Then nth term=84+87=171
171=a + (n-1)d
171=3 + (n-1)x7
171-3/7+1=n
168/7+1=n
24+1=25=n
Therefore 25th term of the ap will be 84 more than 13th term
(8) How many terms of the arithmetic series 24 + 21 + 18 + 15 +g, be taken continuously so that their sum is – 351.
In the given arithmetic series, a = 24, d =- 3.
Let us find n such that Sn = – 351
Now, Sn = n/2[(2a + (n-1)d] 
– 351 = n/2[(48 + (n-1)x(-3)] 
on solving we get,  n2 - 17n -234 = 0
Þ (n - 26h)(n + 9) = 0
Þ  n = 26 or n = - 9
Here n, being the number of terms needed, cannot be negative
Thus, 26 terms are needed to get the sum -351.
(9) Find the sum of the first 2n terms of the following series. 12 - 22 + 32 - 42 +  
We want to find 12 - 22 + 32 - 42 +.....  to  2n terms
= 1 - 4 + 9 - 16 + 25 ------------2n terms
= (1 – 4) +(9 – 16)+(25 – 36) + ----------- to n terms. (after grouping)
= -3 +(-7)+(-11)+ -------------- n terms
Now, the above series is in an A.P. with first term a = - 3 and common difference d = - 4 
Now, Sn = n/2[(2a + (n-1)d]  = = n/2[(2 x -3) + (n-1)(-4)]  = -n(2n + 1).
(10)  A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest-of these angles is 8° and the largest 72°, calculate n and the angle in the fourth sector.
Let the common difference of the A.P. be x
The smallest angle = 8°
 a = 8
And the largest is 72°
 an = 72
 a + (n – 1)d = 72
8 + (n – 1)d = 72
(n – 1) d = 72 – 8 = 64 ... (1)
We know that sum of all the angles of a circle is 360°
Sn = n/2[(2a + (n-1)d]  = 360
Þ Sn = n/2[(2x8 + 64] = 360
Þ n= 9
Putting the value of n in equation (1) we get
(9 – 1) d = 64
d = 8
Now angle in fourth sector = a4 = a + (4 – 1) d
= a + 3d = 8 + 3 × 8 = 8 + 24 = 32

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