ADDITIONAL QUESTIONS:
1. Check whether – 75 is a term of the AP : 21, 18, 15……………… . .
2.Determine the AP whose 3rd term is 5 and the 7th term is 9.
3. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
- Find the 11th term from the last term ( towards the first term ) of the AP: 10, 7, 4, ….. ( − 71 ) .
- Find the 60th term of the AP 8, 10, 12, …, if it has a total of 60 terms.
- In an A.P., if the 12th term is – 13 and the sum of its first four terms is 24, find the sum of its first ten terms.
- How many three digit numbers are divisible by 3.
Geometric Progression:
A Geometric Progression is a sequence of numbers in which we get each term by multiplying or dividing a particular number to the previous term, except the first term.
The ratio between every term to the next term is constant.
EXERCISE 5.3
- Find the sum of the following AP’s.
( i ) 2, 7, 12… to 10 terms ( ii ) – 37, – 33, – 29… to 12 terms
( iii ) 0.6, 1.7, 2.8… to 100 terms ( iv ) 1 / 15, 1 / 12, 1 / 10, …… , to 11 terms
Solution: ( i ) Given, 2, 7, 12 ,…, to 10 terms
For this A.P.,first term, a = 2
common difference, d = a2 − a1 = 7 − 2 = 5 and n = 10
The formula for sum of nth term in AP series is,
Sn = n / 2 [2 a + ( n − 1 ) d]
S10 = 10 / 2 [2 ( 2 ) + ( 10 − 1 ) × 5 ]
= 5 [4 + ( 9 ) × ( 5 ) ]
= 5 × 49 = 245
( ii ) Given, − 37, − 33, − 29 ,…, to 12 terms
For this A.P,first term, a = − 37
common difference, d = a2 − a1
d = ( − 33 ) − ( − 37 )
= − 33 + 37 = 4 and n = 12
The formula for sum of nth term in AP series is,
Sn = n / 2 [2a + ( n − 1 ) d]
S12 = 12 / 2 [2 ( − 37 ) + ( 12 − 1 ) ×4]
= 6 [ − 74 + 11 × 4 ] = 6 [ − 74 + 44]
= 6 (− 30 ) = − 180
( iii ) Given, 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P,the first term, a = 0.6
Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1 and n = 100.
The formula for sum of nth term in AP series is,
Sn = n / 2 [2 a + ( n − 1 ) d ]
S12 = 50 / 2 [1.2 + ( 99 ) ×1.1 ]
= 50 [ 1.2 + 108.9 ]
= 50 [110.1] = 5505 = 5505
( iv ) Given, 1 / 15, 1 / 12, 1 / 10, …… , to 11 terms
For this A.P, First term, a = 1 / 5
Common difference, d = a2 – a1 = ( 1 / 12 ) − − ( 1 / 5 ) = 1 / 60
And number of terms n = 11
The formula for sum of nth term in AP series is,
Sn = n / 2 [ 2a + ( n – 1 ) d ]
= 11 / 2 ( 2 / 15 + 10 / 60 )
= 11 / 2 ( 9 / 30 ) = 33/20.
- Find the sums given below:
( i ) 7 + 10 ( ii ) 34 + 32 + 30 + … + 10
( iii ) – 5 + ( – 8 ) + ( – 11 ) + … + ( – 230 )
Solution: ( i ) First term, a = 7
nth term, an = 84
d = 10 =
Let 84 be the nth term of this A.P., then as per the nth term formula,
an = a ( n − 1 ) d
84 = 7 + ( n – 1 ) × 7 / 2
77 = ( n − 1 ) × 7 / 2
22 = n − 1. \ n = 23
We know that, sum of n term is;
Sn = n / 2 ( a + l ) , l = 84
Sn = 23 / 2 ( 7 + 84 )Sn = ( 23×91 / 2 ) = 2093 / 2
( ii ) Given, 34 + 32 + 30 + ……….. + 10
For this A.P, First term, a = 34
Common difference, d = a2 − a1 = 32 − 34 = − 2
nth term, an = 10
Let 10 be the nth term of this A.P., therefore,
an = a + ( n − 1 ) d
10 = 34 + ( n − 1 ) ( − 2 )
− 24 = ( n − 1 ) ( − 2 ) ®12 = n − 1® n = 13.
We know that, sum of n terms is;
Sn = n / 2 ( a + l ) , l = 10
= 13 / 2 ( 34 + 10 )
= ( 13 × 44 / 2 ) = 13 × 22 = 286.
( iii ) Given, ( − 5 ) + ( − 8 ) + ( − 11 ) + ………… + ( − 230 )
For this A.P, First term, a = − 5
nth term, an = − 230
Common difference, d = a2 − a1 = ( − 8 ) − ( − 5 )
⇒d = − 8 + 5 = − 3
Let − 230 be the nth term of this A.P., and by the nth term formula we know,
an = a + ( n − 1 ) d
− 230 = − 5 + ( n − 1 ) ( − 3 )
− 225 = ( n − 1 ) ( − 3 ) ⇒ ( n − 1 ) = 75 n = 76
And, Sum of n terms,
Sn = n / 2 ( a + l )
= 76 / 2 [ ( − − 5 ) + ( − − 230 ) ] ⇒ 38 ( − 235 ) = − 8930
- In an AP
( i ) Given a = 5, d = 3, an = 50, find n and Sn.
( ii ) Given a = 7, a13 = 35, find d and S13.
( iii ) Given a12 = 37, d = 3, find a and S12.( iv ) Given a3 = 15, S10 = 125, find d and a10.
( v ) Given d = 5, S9 = 75, find a and a9.( vi ) Given a = 2, d = 8, Sn = 90, find n and an.
( vii ) Given a = 8, an = 62, Sn = 210, find n and d.( viii ) Given an = 4, d = 2, Sn = − 14, find n and a
( ix ) Given a = 3, n = 8, S = 192, find d.( x ) Given l = 28, S = 144 and there are total 9 terms. Find a.
Solutions: ( i ) Given that, a = 5, d = 3, an = 50
The formula of the nth term in an AP,
an = a + ( n − 1 ) d,
Therefore, putting the given values, we get,
⇒ 50 = 5 + ( n − 1 ) ×3
⇒ 3 ( n − 1 ) = 45 ⇒ n − 1 = 15 ⇒ n = 16
Now, sum of n terms,
Sn = n / 2 ( a + an )
Sn = 16 / 2 ( 5 + 50 ) = 440
( ii ) Given that, a = 7, a13 = 35
As we know, from the formula of the nth term in an AP,
an = a + ( n − 1 ) d,
Therefore, putting the given values, we get,
⇒ 35 = 7 + ( 13 − 1 ) d ⇒ 12d = 28 ⇒ d = 28 / 12 = 2.33
Now, Sn = n / 2 ( a + an )
S13 = 13 / 2 ( 7 + 35 ) = 273
( iii ) Given that, a12 = 37, d = 3
As we know, from the formula of the nth term in an AP,
an = a + ( n − 1 ) d,
Therefore, putting the given values, we get,
⇒ a12 = a + ( 12 − 1 ) 3 ⇒ 37 = a + 33 ⇒ a = 4
Now, sum of nth term,
Sn = n / 2 ( a + an )
Sn = 12 / 2 ( 4 + 37 ) = 246
( iv ) Given that, a3 = 15, S10 = 125
As we know, from the formula of the nth term in an AP,
an = a + ( n − 1 ) d,
Therefore, putting the given values, we get,
a3 = a + ( 3 − 1 ) d
15 = a + 2d ………………………….. ( i )
Sum of the nth term,
Sn = n / 2 [2 a + ( n − 1 ) d ]
S10 = 10 / 2 [2 a + ( 10 − 1 ) d ]
125 = 5 ( 2 a + 9 d )
25 = 2 a + 9 d ……………………….. ( ii )
On multiplying equation ( i ) by ( ii ) , we will get;
30 = 2 a + 4 d ………………………………. ( iii )
By subtracting equation ( iii ) from ( ii ) , we get,
− 5 = 5 d ⇒ d = − 1
From equation ( i )
15 = a + 2 ( − 1 ) 15 = a − 2
a = 17 = First term
a10 = a + ( 10 − 1 ) d
a10 = 17 + ( 9 ) ( − 1 ) ⇒ 17 − 9 = 8
( v ) Given that, d = 5, S9 = 75
As, sum of n terms in AP is,
Sn = n / 2 [2a + ( n − 1 ) d]
Therefore, the sum of first nine terms are;
S9 = 9 / 2 [ 2 a + ( 9 − 1 ) 5 ]
25 = 3 ( a + 20 ) ⇒ 25 = 3 a + 60
3a = 25 − 60 ⇒ a = − 35 / 3
As we know, the nth term can be written as;
an = a + ( n − 1 ) d
a9 = a + ( 9 − 1 ) ( 5 )
= − 35 / 3 + 8 ( 5 ) ⇒ − 35 / 3 + 40
= ( 35 + 120 / 3 ) ⇒ 85 / 3
( vi ) Given that, a = 2, d = 8, Sn = 90
As, sum of n terms in an AP is,
Sn = n / 2 [2 a + ( n − 1 ) d]
90 = n / 2 [2 a + ( n − 1 ) d]
⇒ 180 = n ( 4 + 8 n − 8 ) = n ( 8 n − 4 ) = 8 n2 − 4 n
⇒ 8 n2 − 4 n – 180 = 0 ⇒ 2 n2 – n − 45 = 0
⇒ 2 n2 − 10 n + 9 n − 45 = 0 ⇒2 n ( n − 5 ) + 9 ( n − 5 ) = 0
⇒ ( n − 5 ) ( 2 n + 9 ) = 0
So, n = 5 ( as n only be a positive integer )
∴ a5 = 8 + 5 × 4 = 34
( vii ) Given that, a = 8, an = 62, Sn = 210
As, sum of n terms in an AP is,
Sn = n / 2 ( a + an )
210 = n / 2 ( 8 + 62 )
⇒ 35 n = 210 ⇒ n = 210 / 35 = 6
Now, 62 = 8 + 5 d
⇒ 5 d = 62 − 8 = 54 ⇒ d = 54 / 5 = 10.8
( viii ) Given that, nth term, an = 4, common difference, d = 2, sum of n terms, Sn = − 14.
As we know, from the formula of the nth term in an AP,
an = a + ( n − 1 ) d,
Therefore, putting the given values, we get,
4 = a + ( n − 1 ) 2 ⇒ 4 = a + 2 n − 2
a + 2 n = 6
a = 6 − 2 n …………………………………………. ( i )
As we know, the sum of n terms is;
Sn = n / 2 ( a + an )
− 14 = n / 2 ( a + 4 )
− 28 = n ( a + 4 )
− 28 = n ( 6 − 2 n + 4 ) {From equation ( i ) }
− 28 = n ( − 2 n + 10 ) − 28 = − 2 n 2 + 10 n
2 n 2 − 10 n − 28 = 0 ⇒ n 2 − 5 n − 14 = 0
n 2 − 7 n + 2 n − 14 = 0 ⇒ n ( n − 7 ) + 2 ( n − 7 ) = 0
( n − 7 ) ( n + 2 ) = 0
Either n − 7 = 0 or n + 2 = 0 ⇒ n = 7 or n = − 2
However, n can neither be negative nor a fraction.
Therefore, n = 7
From equation ( i ) , we get
a = 6 − 2 n
a = 6 − 2 ( 7 ) ⇒ 6 − 14 ⇒ − 8
( ix ) Given that, first term, a = 3,
Number of terms, n = 8 and sum of n terms, S = 192
As we know,
Sn = n / 2 [ 2 a + ( n − 1 ) d ]
192 = 8 / 2 [2 × 3 + ( 8 − 1 ) d ]
192 = 4 [6 + 7d ] ⇒ 48 = 6 + 7d
42 = 7d ⇒ d = 6
( x ) Given that, l = 28, S = 144 and there are total of 9 terms.
Sum of n terms formula,
Sn = n / 2 ( a + l )
144 = 9 / 2 ( a + 28 )
( 16 ) × ( 2 ) = a + 28
32 = a + 28 ⇒ a = 4
- How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636.
Solution: First term = a = 9, Common difference = d = 17 – 9 = 8, Sn = 636
Applying formula, Sn = n / 2 [2 a + ( n − 1 ) d ] to find sum of n terms of AP, we get
636 = [ 18 + ( n − 1 ) ( 8 ) ]
⇒ 1272 = n ( 18 + 8 n − 8 )
⇒ 1272 = 18 n + 8 n2 – 8 n
⇒ 8 n2 + 10 n – 1272 = 0
⇒ 4 n2 + 5 n – 636 = 0
Comparing equation 4 n2 + 5 n – 636 = 0 with general form an2 + bn + c = 0, we get a = 4, b = 5 and c = − 636
We discard negative value of n here because n cannot be in negative, n can only be a positive integer.
Therefore, n = 12
Therefore, 12 terms of the given sequence make sum equal to 636.
5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution: First term = a = 5, Last term = l = 45,
Applying formula, Sn = n / 2 ( a + l ) to find sum of n terms of AP, we get
Applying formula, Sn = n / 2 [2 a + ( n − 1 ) d ] to find sum of n terms of AP and putting value of n, we get
- The first and the last terms of an AP are 17 and 350 respectively. If, the common difference is 9, how many terms are there and what is their sum.
Solution: First term = a = 17, Last term = l = 50 and Common difference = d = 9
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression, we get
350 = 17 + ( n − 1 ) ( 9 )
⇒ 350 = 17 + 9 n − 9
⇒ 342 = 9 n ⇒ n = 38
Applying formula, Sn = n / 2 ( a + l ) to find sum of n terms of AP and putting value of n, we get
S38 = 38 / 2 [ 34 + ( 38 − 1 ) 9 ]
⇒ S38 = 19 ( 34 + 333 ) = 6973
Therefore, there are 38 terms and their sum is equal to 6973.
- Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:It is given that 22nd term is equal to 149 / a 22 = 149
Using formula an = a + ( n − 1 ) d , to find nth term of arithmetic progression, we get
149 = a + ( 22 − 1 ) ( 7 )
⇒ 149 = a + 147 ⇒ a = 2
Applying formula, Sn = n / 2 [ 2a + ( n − 1 ) d ] to find sum of n terms of AP and putting value of a, we get
S22 = 22 / 2 [4 + ( 22 − 1 ) 7 ]
⇒ S22 = 11 ( 4 + 147 )
⇒ S22 = 1661
Therefore, sum of first 22 terms of AP is equal to 1661.
- Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution: It is given that second and third term of AP are 14 and 18 respectively.
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression, we get
14 = a + ( 2 − 1 ) d
⇒ 14 = a + d … ( 1 )
And, 18 = a + ( 3 − 1 ) d
⇒ 18 = a + 2 d … ( 2 )
These are equations consisting of two variables.
Using equation ( 1 ) , we get, a = 14 − d
Putting value of a in equation ( 2 ) , we get
18 = 14 – d + 2 d
⇒ d = 4
Therefore, common difference d = 4
Putting value of d in equation ( 1 ) , we get
18 = a + 2 ( 4 )
⇒ a = 10
Applying formula, Sn = n / 2 [ 2 a + ( n – 1 ) d ] to find sum of n terms of the AP, we get
S51 = 51 / 2 [ 2 × 10 ( 51 − 1 ) 4]
= 51 / 2 [ 2 + ( 20 ) × 4]
= 51 × 220 / 2 = 51 × 110 = 5610
Therefore, sum of first 51 terms of an AP is equal to 5610.
- If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution: It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289.
Applying formula, Sn = n / 2 [ 2a + ( n – 1 ) d ] to find sum of n terms of AP, we get
49 = 7 / 2 [ 2 a + ( 7 − 1 ) d]
⇒ 98 = 7 ( 2 a + 6 d )
⇒ 7 = a + 3 d ⇒ a = 7 − 3 d … ( 1 )
289 = 17 / 2 ( 2 a + 16 d )
⇒ 578 = 17 ( 2 a + 16 d ) ⇒ 34 = 2 a + 16 d ⇒ 17 = a + 8 d
Putting equation ( 1 ) in the above equation, we get
17 = 7 − 3 d + 8 d
⇒ 10 = 5 d ⇒ d = 2
Putting value of d in equation ( 1 ) , we get
a = 7 − 3 d = 7 – 3 ( 2 ) = 7 – 6 = 1
Again applying formula, Sn = n / 2 [ 2 a + ( n − 1 ) d ] to find sum of n terms of AP, we get
= n / 2 [2 ( 1 ) + ( n – 1 ) ×2]
= n / 2 ( 2 + 2 n − 2 )
= n / 2 ( 2 n )
Sn = n2
Therefore, sum of n terms of AP is equal to n2.
- Show that a1, a2 … , an, … form an AP where an is defined as below
( i ) an = 3 + 4 n
( ii ) an = 9 − 5 n
Also find the sum of the first 15 terms in each case.Solution: ( i ) an = 3 + 4 n
a1 = 3 + 4 ( 1 ) = 7, a2 = 3 + 4 ( 2 ) = 3 + 8 = 11
a3 = 3 + 4 ( 3 ) = 3 + 12 = 15 a4 = 3 + 4 ( 4 ) = 3 + 16 = 19
We can see here, the common difference between the terms are;
a2 − a1 = 11 − 7 = 4 a3 − a2 = 15 − 11 = 4 a4 − a3 = 19 − 15 = 4
Hence, ak + 1 − ak is the same value every time. ∴ This is an AP with common difference as 4 and first term as 7.
Now, we know, the sum of nth term is;
Sn = n / 2 [ 2a + ( n − 1 ) d ]
S15 = 15 / 2 [ 2 ( 7 ) + ( 15 − 1 ) ×4 ]
= 15 / 2 [ ( 14 ) + 56 ] = 15 / 2 ( 70 ) = 15 × 35 = 525
( ii ) an = 9 − 5 n
a1 = 9 − 5×1 = 9 − 5 = 4 a2 = 9 − 5×2 = 9 − 10 = − 1
a3 = 9 − 5×3 = 9 − 15 = − 6 a4 = 9 − 5×4 = 9 − 20 = − 11
We can see here, the common difference between the terms are;
a2 − a1 = − 1 − 4 = − 5 a3 − a2 = − 6 − ( − 1 ) = − 5 and so on
Hence, ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as − 5 and first term as 4. Now, we know, the sum of nth term is;
Sn = n / 2 [ 2 a + ( n − 1 ) d ]
S15 = 15 / 2[ 2 ( 4 ) + ( 15 −1 ) ( −5 ) ]
= 15 / 2[8 + 14 ( −5 ) ] = 15 / 2 ( 8−70 )
= 15 / 2 ( −62 ) = 15 ( −31 ) = −465
- If the sum of the first n terms of an AP is 4n − n2, what is the first term ( that is S1 ) What is the sum of first two terms. What is the second term. Similarly, find the 3rd, the 10th and the nth terms.
. It is given that the sum of n terms of an AP is equal to 4n − n2
It means Sn = 4n − n2
Let us calculate S1 and S2 using Sn = 4n − n2
S1 = 4 ( 1 ) − ( 1 ) 2 = 4 – 1 = 3
S2 = 4 ( 2 ) − ( 2 ) 2 = 8 – 4 = 4
First term = a = S1 = 3 … ( 1 )
Let us find common difference now
We can write any AP in the form of general terms like a , a + d, a + 2d …
We have calculated that sum of first two terms is equal to 4 ∴ S2 = 4
Therefore, we can say that a + ( a + d ) = 4
Putting value of a from equation ( 1 ) , we get
2 a + d = 4 ⇒ 2 ( 3 ) + d = 4 ⇒ 6 + d = 4 ⇒ d = – 2
Using formula an = a + ( n – 1 ) d, to find nth term of arithmetic progression,
Second term of AP = a2 = a + ( 2 – 1 ) d = 3 + ( 2 – 1 ) ( – 2 ) = 3 – 2 = 1
Third term of AP = a3 = a + ( 3 – 1 ) d = 3 + ( 3 – 1 ) ( – 2 ) = 3 – 4 = – 1
Tenth term of AP = a10 = a + ( 10 – 1 ) d = 3 + ( 10 – 1 ) ( – 2 ) = 3 – 18 = – 15
nth term of AP = an = a + ( n – 1 ) and d = 3 + ( n – 1 ) ( – 2 ) = 3 – 2n + 2 = 5 – 2n
- Find the sum of the first 40 positive integers divisible by 6.
Solution: The first 40 positive integers divisible by 6 are 6, 12, 18, 24 … 40 terms.
Therefore, we want to find sum of 40 terms of sequence of the form:
6, 12, 18, 24 … 40 terms
Here, first term = a = 6 and Common difference = d = 12 – 6 = 6, n = 40
Applying formula, Sn = n / 2 [ 2a + ( n – 1 ) d ] to find sum of n terms of AP, we get
S40 = 40 / 2 [2 ( 6 ) + ( 40 − 1 ) 6 ]
= 20 [12 + ( 39 ) ( 6 ) ]
= 20 ( 12 + 234 ) = 20 × 246 = 4920
13. Find the sum of the first 15 multiples of 8.
Solution:The multiples of 8 are 8, 16, 24, 32…
The series is in the form of AP, having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8 ∴ S15 = ?
By the formula of sum of nth term, we know,
Sn = n / 2 [ 2a + ( n − 1 ) d ]
S15 = 15 / 2 [2 ( 8 ) + ( 15 − 1 ) 8 ]
= 15 / 2 [ 6 + ( 14 ) ( 8 ) ]
= 15 / 2 [16 + 112 ] = 15 ( 128 ) / 2 = 15 × 64 = 960
- Find the sum of the odd numbers between 0 and 50.
Solution:The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.
Therefore, we can see that these odd numbers are in the form of A.P.
Hence, First term, a = 1 Common difference, d = 2 Last term, l = 49
By the formula of last term, we know,
l = a + ( n − 1 ) d
49 = 1 + ( n − 1 ) 2
48 = 2 ( n − 1 )
n − 1 = 24 n = 25 = Number of terms
By the formula of sum of nth term, we know,
Sn = n / 2 ( a + l )
S25 = 25 / 2 ( 1 + 49 )
= 25 ( 50 ) / 2 = ( 25 ) ( 25 ) = 625
- A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
Solution: Penalty for first day = Rs 200, Penalty for second day = Rs 250
Penalty for third day = Rs 300
It is given that penalty for each succeeding day is Rs 50 more than the preceding day.
It makes it an arithmetic progression because the difference between consecutive terms is constant.
We want to know how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
So, we have an AP of the form 200, 250, 300, 350 … 30 terms
First term = a = 200, Common difference = d = 50, n = 30
Applying formula, Sn = n / 2 [2a + ( n − 1 ) d ] to find sum of n terms of AP, we get
S30 = 30 / 2 [2 ( 200 ) + ( 30 – 1 ) 50 ]
= 15 [ 400 + 1450 ]
= 15 ( 1850 ) = 27750
Therefore, penalty for 30 days is Rs. 27750.
- A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is Rs 20 less than its preceding term, find the value of each of the prizes.
Solution: It is given that sum of seven cash prizes is equal to Rs 700.
And, each prize is R.s 20 less than its preceding term.
Let value of first prize = Rs. a
Let value of second prize = Rs ( a − 20 )
Let value of third prize = Rs ( a − 40 )
So, we have sequence of the form:
a, a − 20, a − 40, a – 60 …
It is an arithmetic progression because the difference between consecutive terms is constant.
First term = a, Common difference = d = ( a – 20 ) – a = – 20
n = 7 ( Because there are total of seven prizes )
S7 = 700{given}
Applying formula, Sn = n / 2 [ 2a + ( n − 1 ) d ] to find sum of n terms of AP, we get
7 / 2 [ 2 a + ( 7 – 1 ) d ] = 700
⇒ 200 = 2 a – 120
⇒ 320 = 2 a ⇒ a = 160
Therefore, value of first prize = Rs 160
Value of second prize = 160 – 20 = Rs 140
Value of third prize = 140 – 20 = Rs 120
Value of fourth prize = 120 – 20 = Rs 100
Value of fifth prize = 100 – 20 = Rs 80
Value of sixth prize = 80 – 20 = Rs 60
Value of seventh prize = 60 – 20 = Rs 40
- In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g, a section of Class I will plant 1 tree, a section of class II will plant two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students.
Solution: There are three sections of each class and it is given that the number of trees planted by any class is equal to class number. x 2 = 3 x 2 = 6
The number of trees planted by class I = number of sections x 1 = 3 x 1 = 3
The number of trees planted by class II = number of sections x 2 = 3 x 2 = 6
The number of trees planted by class III = number of sections x 3 = 3 x 3 = 9
Therefore, we have sequence of the form 3, 6, 9 … 12 terms
To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 … 12 terms.
First term a = 3, Common difference d = 6 – 3 = 3 and n = 12
Applying formula, to find sum of n terms of AP , we get
Sn = n / 2 [ 2a + ( n − 1 ) d ]
S12 = 12 / 2 [ 2 ( 1 ) + ( 12 − 1 ) ( 1 ) ]
= 6 ( 2 + 11 ) = 6 ( 13 ) = 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3×78 = 234
Therefore, 234 trees will be planted by the students.
- A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … What is the total length of such a spiral made up of thirteen consecutive semicircles.
Solution:We know, Perimeter of a semi circle = πr
P1 = π ( 0.5 ) = π / 2 cm
P2 = π ( 1 ) = π cm
P3 = π ( 1.5 ) = 3π / 2 cm where P1 , P2 , P3 are the lengths of the semi circles.
Hence we got a series here, as, π / 2, π, 3π / 2, 2π, ….
P1 = π / 2 cm
P2 = π cm
Common difference, d = P2 – P1 = π – π / 2 = π / 2
First term = P1 = a = π / 2 cm
By the sum of n term formula, we know,
Sn = n / 2 [ 2a + ( n – 1 ) d]
∴ sum of the length of 13 consecutive circles is;
S13 = 13 / 2 [ 2 ( π / 2 ) + ( 13 – 1 ) π / 2 ]
= 13 / 2 [π + 6 π ]
= 13 / 2 ( 7 π ) = 13 / 2 × 7 × 22 / 7 = 143 cm
- 200 logs are stacked in the following manner:20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row.
Solution: The number of logs in the bottom row = 20
The number of logs in the next row = 19
The number of logs in the next to next row = 18
Therefore, we have sequence of the form 20, 19, 18 …
First term a = 20, Common difference d = 19 – 20 = – 1
We need to find that how many rows make total of 200 logs.
By the sum of nth term formula,
Sn = n / 2 [ 2a + ( n − 1 ) d ]
S12 = 12 / 2 [2 ( 20 ) + ( n − 1 ) ( − 1 ) ]
400 = n ( 40 − n + 1 ) ⇒ 400 = n ( 41 − n ) ⇒ 400 = 41n − n2
∴ n2 − 41n + 400 = 0
solving this equation we get n2 − 16n − 25n + 400 = 0
n ( n − 16 ) − 25 ( n − 16 ) = 0 ( n − 16 ) ( n − 25 ) = 0
Either ( n − 16 ) = 0 or n − 25 = 0
n = 16 or n = 25
By the nth term formula,
an = a + ( n − 1 ) d ⇒ a16 = 20 + ( 16 − 1 ) (− 1 )
a16 = 20 − 15 ⇒ a16 = 5
Similarly, the 25th term could be written as;
a25 = 20 + ( 25 − 1 ) ( − 1 ) ⇒ a25 = 20 − 24 = − 4
It can be seen, the number of logs in 16th row is 5 as the numbers cannot be negative.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.
- In a potato race, a bucket is placed at the starting point, which is 5 meters from the first potato, and the other potatoes are placed 3 meters apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run
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Solution:The distance of first potato from the starting point = 5 meters
Therefore, the distance covered by competitor to pick up first potato and put it in bucket = 5 x 2 = 10 meters
The distance of Second potato from the starting point = 5 + 3 = 8 meters {All the potatoes are 3 meters apart from each other}
Therefore, the distance covered by competitor to pick up 2nd potato and put it in bucket = 8 x 2 = 16 meters
The distance of third potato from the starting point = 8 + 3 = 11 meters
Therefore, the distance covered by competitor to pick up 3rd potato and put it in bucket = 11 x 2 = 22 meters
Therefore, we have a sequence of the form 10, 16, 22 … 10 terms
( There are ten terms because there are ten potatoes )
To calculate the total distance covered by the competitor, we need to find:
10 + 16 + 22 + … 10 terms
First term = a = 10, Common difference = d = 16 – 10 = 6
n = 10{There are total of 10 terms in the sequence}
S10 = 12 / 2 [ 2 ( 20 ) + ( n − 1 ) ( − 1 ) ]
= 5 [ 20 + 54 ] = 5 ( 74 ) = 370
Therefore, the competitor will run a total distance of 370 m.
Additional Questions:
1.Find the sum of the first 13 multiples of 7.
2.Find the sum of the odd numbers between 1 and 61.
3.A sum of Rs 1000 is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is Rs 50 less than its preceding term, find the value of each of the prizes.
- The first and last term of an AP are 4 and 81 respectively. if the common difference is 7, how many terms are there in the AP and what is their sum.
- The sum of three numbers in A.P. is 12 and sum of their cubes is 288, Find the numbers.
6.The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses proceding the house numbered X is equal to sum of the numbers of houses following X.