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
a (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
,
a (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
∴ The value of n = 9 and angle in fourth sector is 32°
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10th Maths SA-2 Chapter Links
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