5.4 SUM OF FIRST N TERMS OF AN ARITHMETIC SERIES:
Sum of the first n terms of the sequence is calculated by
an = a + ( n – 1 ) d.
Example 1 :Find the 10th term of the AP : 2, 7, 12, . . .
Solution : Here, a = 2, d = 7 – 2 = 5 and n = 10.
We have an = a + ( n – 1 ) d
So, a = 2 + ( 10 – 1 ) × 5 = 2 + 45 = 47
Therefore, the 10th term of the given AP is 47.
Example 2: Which term of the AP: 21, 18, 15, …is – 81. Also,is any term 0. Give reasons for your answer.
Solution : Here, a = 21, d = 18 – 21 = – 3 and an = – 81, and we have to find n.
As an = a + ( n – 1 ) d,
we have – 81 = 21 + ( n – 1 ) ( – 3 )
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35
Therefore, the 35th term of the given AP is – 81.
Also if there is any n for which an = 0. If such an n is there,
then 21 + ( n – 1 ) ( – 3 ) = 0,
3 ( n – 1 ) = 21
n = 8
So, the eighth term is 0.
Example 3 : Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution : We have an = a + ( n – 1 ) d,
a3 = a + ( 3 – 1 ) d = a + 2 d = 5 ( 1 ) and
a7 = a + ( 7 – 1 ) d = a + 6 d = 9 ( 2 )
Solving the pair of linear equations ( 1 ) and ( 2 ) , we get
a = 3 and d = 1
Hence, the required AP is 3, 4, 5, 6, 7, .
Example 4: Check whether 301 is a term of the list of numbers
5, 11, 17, 23, . . ……….
Solution : We have : an = a + ( n – 1 ) d,
( a2 – a1) = 11 − 5 = 6 and ( a3 – a2 ) = 17 – 11 = 6, and so on
Since ak + 1 – ak is the same for k = 1, 2, 3, etc., the given list of numbers is an AP. Now, a = 5 and d = 6.
Let 301 be a term, say, the nth term of this AP. We know that
an = a + ( n – 1 ) d
So, 301 = 5 + ( n – 1 ) × 6
301 = 6n – 1
But n should be a positive integer . So, 301 is not a term of the given list of numbers.
Example 5 :How many two digit numbers are divisible by 3.
Solution :The list of two digit numbers divisible by 3 is :
12, 15, 18, . . . , 99
Here, a = 12, d = 3, an = 99.
an = a + ( n – 1 ) d,
We have 99 = 12 + ( n – 1 ) × 3
87 = ( n – 1 ) × 3
87 = 3n − 3 ⇒ 90 = 3n ⇒ n = 30.
The number of two digit numbers divisible by 3 are 30
Example 6 : Find the 11th term from the last term ( towards the first term ) of the AP : 10, 7, 4, . . ., – 62.
Solution: Here, a = 10, d = 7 – 10 = – 3, l = – 62,
where l = a + ( n – 1 ) d ( l = an )
We will find the total number of terms in the AP.
So, – 62 = 10 + ( n – 1 ) ( – 3 )
– 72 = ( n – 1 ) ( – 3 )
n – 1 = 24
or n = 25
So, there are 25 terms in the given AP.
The 11th term from the last term will be the 15th term.
So, a = 10 + ( 15 – 1 ) ( – 3 ) = 10 – 42 = – 32
the 11th term from the last term is – 32.
Example 7: A sum of`1000 is invested at 8% simple interest per year.Calculate the interest at the end of each year. Do these interests form an AP. If so,find the interest at the end of 30 years making use of this fact.
Solution :We know that the formula to calculate simple interest is given by
Similarly we can obtain the interest at the end of the 4th year, 5th year, and so on. So, the interest ( in Rs ) at the end of the 1st, 2nd, 3rd, . . . years, respectively are 80, 160, 240 ,
It is an AP as the difference between the consecutive terms in the list is 80
d = 80. Also, a = 80.
EXERCISE 5.2
- Find the missing variable from a, d, n and an, where a is the first term, d is the common difference and an is the nth term of AP.
( i ) a = 7, d = 3, n = 8
( ii ) a = – 18, n = 10,
( iii ) d = – 3, n = 18,
( iv ) a = – 18.9, d = 2.5,
( v ) a = 3.5, d = 0, n = 105
Solution: ( i ) a = 7, d = 3, n = 8
We need to find an here.
Using formula an = a + ( n – 1 ) d
Putting values of a, d and n,
an = 7 + ( 8 – 1 ) 3
= 7 + ( 7 ) 3 = 7 + 21 = 28
( ii ) Solution: a = – 18, n = 10,
We need to find d here.
Using formula = a + ( n – 1 ) d
Putting values of a, and n,
0 = – 18 + ( 10 – 1 ) d
⇒ 0 = − 18 + 9 d
⇒ 18 = 9 d ⇒ d = 2
( iii ) Solution: d = – 3, n = 18,
We need to find a here.
Using formula an = a + ( n – 1 ) d
Putting values of of a , d and n,
– 5 = a + ( 18 – 1 ) ( – 3 )
⇒ − 5 = a + ( 17 ) ( − 3 )
⇒ − 5 = a – 51 ⇒ a = 46
( iv ) Solution: a = – 18.9, d = 2.5
We need to find n here.
Using formula an = a + ( n – 1 ) d
Putting values of d, an and n ,
3.6 = – 18.9 + ( n – 1 ) ( 2.5 )
⇒ 3.6 = − 18.9 + 2.5 n − 2.5
⇒ 2.5 n = 25 ⇒ n = 10
( v ) Solution: a = 3.5 d = 0, n = 105
We need to find an here.
Using formula an = a + ( n – 1 ) d
Putting values of d, n and a,
an = 3.5 + ( 105 − 1 ) ( 0 )
an = 3.5 + ( 105 − 1 ) ( 0 )
= 3.5 + ( 105 − 1 ) ( 0 )
= 3.5 + 104 x 0
= 3.5
an = 3.5
2.Choose the correct choice in the following and justify:
( i ) 30th term of the AP: 10, 7, 4… is
( A ) 97 ( B ) 77 ( C ) – 77 ( D ) – 87
( ii ) 11th term of the AP: − 3, − ½, 2… is
( A ) 28 ( B ) 22 ( C ) – 38 ( D ) − 4 8
Solution: ( i ) 10, 7, 4…
First term = a = 10, Common difference = d = 7 – 10 = 4 – 7 = – 3
And n = 30 (Because, we need to find 30th term)
an = a + ( n – 1 ) d
⇒ a30 = 10 + ( 30 − 1 ) ( − 3 ) = 10 – 87 = − 77
Therefore, the answer is ( C ) .
( ii ) Solution: − 3, − ½, 2…
First term = a = – 3, Common difference = d = − ½ − (− 3 ) = 5 /2 − =
And n = 11 ( Because, we need to find 11th term )
an = − 3 + ( 11 – 1 ) x 5/2 = − 3 + 25 = 22
Therefore 11th term is 22 which means answer is ( B ) .
- In the following AP’s find the missing terms:
( i ) 2,__ , 26 ( ii ) __, 13, __, 3 ( iii ) 5, __, __, 9
( iv ) – 4. __, __, __, __, 6 ( v ) __, 38, __, __, __, – 22
Solution: ( i ) For the given A.P., 2 , _, 26.
The first and third term are;
a1 = 2
a3 = 26
We know, for an A.P., an = a + ( n − 1 ) d
Therefore, putting the values here,
a3 = 2 + ( 3 − 1 ) d
26 = 2 + 2d ⇒ 24 = 2d ∴ d = 12
a2 = 2 + ( 2 − 1 ) 12
∴ a2 = 14
Therefore, 14 is the missing term.
( ii ) For the given A.P. _, 13,_ ,3
a2 = 13 and
a4 = 3
As we know, for an A.P.,
an = a + ( n − 1 ) d
Therefore, putting the values here,
a2 = a + ( 2 − 1 ) d
13 = a + d ………………. ( i )
a4 = a + ( 4 − 1 ) d
3 = a + 3d ………….. ( ii )
On subtracting equation ( i ) from ( ii ) , we get,
– 10 = 2d
d = – 5
From equation ( i ) , putting the value of d,we get
13 = a + ( − 5 )
a1 = 18
a3 = 18 + ( 3 − 1 ) ( − 5 )
= 18 + 2 ( − 5 ) = 18 − 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
( iii ) For the given A.P., 5, _ , _ 9
a1 = 5 and a4 = 19 / 2
As we know, for an A.P.,
an = a + ( n − 1 ) d
Therefore, putting the values here,
a4 = a + ( 4 − 1 ) d
19 / 2 = 5 + 3 d
( 19 / 2 ) – 5 = 3 d ⇒ 3 d = 9 / 2
d = 3 / 2
a2 = a + ( 2 − 1 ) d
a2 = 5 + 3 / 2 ⇒ a2 = 13 / 2
a3 = a + ( 3 − 1 ) d ⇒ a3 = 5 + 2×3 / 2
a3 = 8
Therefore, the missing terms are 3 / 2 and 8 respectively.
( iv ) For the given A.P. – 4. __, __, __, __, 6
a = − 4 and a6 = 6
As we know, for an A.P.,
an = a + ( n − 1 ) d
Therefore, putting the values here,
a6 = a + ( 6 − 1 ) d
6 = − 4 + 5 d
10 = 5 d ⇒ d = 2
a2 = a + d = − 4 + 2 = − 2
a3 = a + 2 d = − 4 + 2 ( 2 ) = 0
a4 = a + 3 d = − 4 + 3 ( 2 ) = 2
a5 = a + 4 d = − 4 + 4 ( 2 ) = 4
Therefore, the missing terms are − 2, 0, 2, and 4 respectively.
( v ) For the given A.P.__, 38, __, __, __, – 22
a2 = 38 and a6 = − 22
As we know, for an A.P.,
an = a + ( n − 1 ) d
Therefore, putting the values here,
a2 = a + ( 2 − 1 ) d
38 = a + d ……………………. ( i )
a6 = a + ( 6 − 1 ) d
− 22 = a + 5 d …………………. ( ii )
On subtracting equation ( i ) from ( ii ) , we get
− 22 − 38 = 4 d
− 60 = 4 d
d = − 15
a = a2 − d = 38 − ( − 15 ) = 53
a3 = a + 2 d = 53 + 2 ( − 15 ) = 23
a4 = a + 3 d = 53 + 3 ( − 15 ) = 8
a5 = a + 4 d = 53 + 4 ( − 15 ) = − 7
Therefore, the missing terms are 53, 23, 8, and − 7 respectively.
- Which term of the AP: 3, 8, 13, 18 … is 78.
Solution: Given the A.P. series as 3, 8, 13, 18, …
First term, a = 3
Common difference, d = a2 − a1 = 8 − 3 = 5
Let the nth term of given A.P. be 78. Now as we know,
an = a + ( n − 1 ) d
∴ 78 = 3 + ( n − 1 ) 5
75 = ( n − 1 ) 5 ⇒ ( n − 1 ) = 15
n = 16
Hence, 16th term of this A.P. is 78.
- Find the number of terms in each of the following APs:
i ) 7, 13, 19, …, 205
( ii ) 18, 15 ,13….. − 47
Solution: ( i ) Given, 7, 13, 19, …, 205 is the A.P
Therefore First term, a = 7
Common difference, d = a2 − a1 = 13 − 7 = 6
Let there are n terms in this A.P.
an = 205
As we know, for an A.P.,
an = a + ( n − 1 ) d
Therefore, 205 = 7 + ( n − 1 ) 6
198 = ( n − 1 ) 6
33 = ( n − 1 )
n = 34
Therefore, this given series has 34 terms in it.
( ii ) Given 18, 15 ,13….. − 47 is the A.P
First term, a = 18
Common difference, d = a2 − a1 = 15
d = ( 31 − 36 ) / 2 = − 5 / 2
Let there are n terms in this A.P.
an = 205
As we know, for an A.P.,
an = a + ( n − 1 ) d
− 47 = 18 + ( n − 1 ) ( − 5 / 2 )
− 47 − 18 = ( n − 1 ) ( − 5 / 2 )
− 65 = ( n − 1 ) ( − 5 / 2 )
( n − 1 ) = − 130 / − 5
( n − 1 ) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.
- Check whether – 150 is a term of the AP: 11, 8, 5, 2…
Solution: For the given series, A.P. 11, 8, 5, 2..
First term, a = 11
Common difference, d = a2 − a1 = 8 − 11 = − 3
Let − 150 be the nth term of this A.P.
As we know, for an A.P.,
an = a + ( n − 1 ) d
− 150 = 11 + ( n − 1 ) ( − 3 )
− 150 = 11 − 3n + 3
− 164 = − 3n
n = 164 / 3
Clearly, n is not an integer but a fraction.
Therefore, – 150 is not a term of this A.P.
- Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution:Given that, 11th term, a11 = 38
and 16th term, a16 = 73
We know that,
an = a + ( n − 1 ) d
∴ a11 = a + ( 11 − 1 ) d
38 = a + ( 11 − 1 ) ( d ) and 73 = a + ( 16 − 1 ) ( d )
⇒ 38 = a + 10 d and 73 = a + 15 d
These are equations consisting of two variables.
We have, 38 = a + 10 d
⇒ a = 38 − 10 d
Let us put value of a in equation ( 73 = a + 15 d ) ,
73 = 38 − 10 d + 15 d
⇒ 35 = 5 d
∴ common difference = d = 7
Putting value of d in equation 38 = a + 10 d,
38 = a + 70
⇒ a = − 32
Therefore, common difference = d = 7 and First term = a = – 32
Using formula an = a + ( n − 1 ) d
, to find nth term of arithmetic progression,
= − 32 + ( 31 − 1 ) ( 7 )
= − 32 + 210 = 178
∴ 31st term of AP is 178.
8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Ans. An AP consists of 50 terms and the 50th term is equal to 106 and 3rd term = 12
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
= a + ( 50 − 1 ) d and = a + ( 3 − 1 ) d
⇒ 106 = a + 49 d and 12 = a + 2d
These are equations consisting of two variables.
Using equation 106 = a + 49 d, we get a = 106 − 49 d
Putting value of a in the equation 12 = a + 2 d,
12 = 106 − 49 d + 2 d
⇒ 47 d = 94 ⇒ d = 2
Putting value of d in the equation, a = 106 − 49 d,
a = 106 – 49 ( 2 ) = 106 – 98 = 8
Therefore, First term = a = 8 and common difference = d = 2
To find 29th term, we use formula an = a + ( n − 1 ) d
which is used to find nth term of arithmetic progression,
= 8 + ( 29 − 1 ) 2 = 8 + 56 = 64
Therefore, 29th term of AP is equal to 64.
- If the third and the ninth terms of an AP are 4 and – 8 respectively, which term of this AP is zero.
Solution: Given that, 3rd term, a3 = 4
and 9th term, a9 = − 8
We know that, an = a + ( n − 1 ) d
∴ a3 = a + ( 3 − 1 ) d
4 = a + 2d ……………………………………… ( i )
a9 = a + ( 9 − 1 ) d
− 8 = a + 8d ………………………………………………… ( ii )
On subtracting equation ( i ) from ( ii ) , we will get here,
− 12 = 6d
d = − 2
From equation ( i ) , we can write,
4 = a + 2 ( − 2 )
4 = a − 4
a = 8
Let nth term of this A.P. be zero.
an = a + ( n − 1 ) d
0 = 8 + ( n − 1 ) ( − 2 )
0 = 8 − 2n + 2
2n = 10
n = 5
Hence, 5th term of this A.P. is 0.
- The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution: We know that, for an A.P series;
an = a + ( n − 1 ) d
a17 = a + ( 17 − 1 ) d
a17 = a + 16d
In the same way,
a10 = a + 9d
As it is given in the question,
a17 − a10 = 7
( a + 16d ) − ( a + 9d ) = 7
7d = 7
d = 1
∴ the common difference is 1.
- Which term of the AP: 3, 15, 27, 39… will be 132 more than its 54th term.
Solution: Given A.P. is 3, 15, 27, 39, …
first term, a = 3
common difference, d = a2 − a1 = 15 − 3 = 12
We know that, an = a + ( n − 1 ) d
a54 = a + ( 54 − 1 ) d
⇒ 3 + ( 53 ) ( 12 )
⇒ 3 + 636 = 639
a54 = 639
We have to find the term of this A.P. which is 132 more than a54, i.e .771.
Let nth term be 771.
an = a + ( n − 1 ) d
771 = 3 + ( n − 1 ) 12
768 = ( n − 1 ) 12
( n − 1 ) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.
Or another method is;
Let nth term be 132 more than 54th term.
n = 54 + 132 / 2
= 54 + 11 = 65th term
- Two AP’s have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms.
Solution: Let first term of 1st AP = a
Let first term of 2nd AP = a′
It is given that their common difference is same.
Let their common difference be d.
It is given that difference between their 100th terms is 100.
Using formula an = a + ( n − 1 ) d
, to find nth term of arithmetic progression,
a + ( 100 − 1 ) d – [a′ + ( 100 − 1 ) d]
= a + 99 d − a′ − 99 d = 100
⇒ a − a′ = 100… ( 1 )
We want to find difference between their 1000th terms which means we want to calculate:
a + ( 1000 − 1 ) d – [a′ + ( 1000 − 1 ) d]
= a + 999 d − a′ − 999 d = a – a′
Putting equation ( 1 ) in the above equation,
a + ( 1000 − 1 ) d – [a′ + ( 1000 − 1 ) d]
= a + 999 d − a′ + 999 d = a − a′ = 100
Therefore, difference between their 1000th terms would be equal to 100.
- How many three digit numbers are divisible by 7.
Solution: We have AP starting from 105 because it is the first three digit number divisible by 7.
AP will end at 994 because it is the last three digit number divisible by 7.
Therefore, we have AP of the form 105, 112, 119…, 994
Let 994 is the nth term of AP.
We need to find n here.
First term = a = 105, Common difference = d = 112 – 105 = 7
Using formula an = a + ( n − 1 ) d
, to find nth term of arithmetic progression,
994 = 105 + ( n − 1 ) ( 7 )
⇒ 994 = 105 + 7n − 7
⇒ 896 = 7n ⇒ n = 128
It means 994 is the 128th term of AP.
∴ there are 128 terms in AP.
- How many multiples of 4 lie between 10 and 250.
Solution: First multiple of 4 which lie between 10 and 250 is 12.
The last multiple of 4 which lie between 10 and 250 is 248.
Therefore, AP is of the form 12, 16, 20… ,248
First term = a = 12, Common difference = d = 4
Using formula an = a + ( n – 1 ) d, to find nth term of arithmetic progression,
248 = 12 + ( n − 1 ) ( 4 )
⇒ 248 = 12 + 4n − 4
⇒ 240 = 4n ⇒ n = 60
It means that 248 is the 60th term of AP.
So, we can say that there are 60 multiples of 4 which lie between 10 and 250.
- For what value of n, are the nth terms of two AP’s: 63, 65, 67… and 3, 10, 17… equal.
Solution: Lets first consider AP 63, 65, 67…
First term = a = 63, Common difference = d = 65 – 63 = 2
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
an = 63 + ( n − 1 ) ( 2 ) … ( 1 )
Now, consider second AP 3, 10, 17…
First term = a = 3, Common difference = d = 10 – 3 = 7
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
an = 3 + ( n − 1 ) ( 7 ) … ( 2 )
According to the given condition:
( 1 ) = ( 2 )
⇒ 63 + ( n − 1 ) ( 2 ) = 3 + ( n − 1 ) ( 7 )
⇒ 63 + 2n – 2 = 3 + 7n − 7
⇒ 65 = 5n⇒ n = 13
Therefore, 13th terms of both the AP’s are equal.
- Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution: Let first term of AP = a
Let common difference of AP = d
It is given that its 3rd term is equal to 16.
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
16 = a + ( 3 − 1 ) ( d )
⇒ 16 = a + 2d… ( 1 )
It is also given that 7th term exceeds 5th term by 12.
According to the given condition:
⇒ a + ( 7 – 1 ) d = a + ( 5 – 1 ) d + 12
⇒ 2d = 12 ⇒ d = 6
Putting value of d in equation 16 = a + 2d,
16 = a + 2 ( 6 ) ⇒ a = 4
Therefore, first term = a = 4
And, common difference = d = 6
Therefore, AP is 4, 10, 16, 22…
- Find the 20th term from the last term of the AP: 3, 8, 13… , 253.
Solution: We want to find 20th term from the last term of given AP.
So, let us write given AP in this way:253 … 13, 8, 3
Here First term = a = 253, Common Difference = d = 8 – 13 = – 5
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
= 253 + ( 20 − 1 ) ( − 5 )
⇒ = 253 + 19 ( − 5 ) = 253 – 95 = 158
Therefore, the 20th term from the last term of given AP is 158.
- The sum of the 4th and 8th terms of an AP is 24 and the sum of 6th and 10th terms is 44. Find the three terms of the AP.
Solution: The sum of 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44.
and
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
a + ( 4 − 1 ) d + [a + ( 8 − 1 ) d] = 24
And, a + ( 6 − 1 ) d + [a + ( 10 − 1 ) d] = 44
⇒ a + 3 d + a + 7 d
= 24 And a + 5 d + a + 9 d = 44
⇒ 2a + 10 d = 24 and 2 a + 14 d = 44
⇒ a + 5d = 12 And a + 7 d = 22
These are equations in two variables.
Using equation, a + 5d = 12, we can say that a = 12 − 5d… ( 1 )
Putting ( 1 ) in equation a + 7d = 22,
12 − 5 d + 7 d = 22
⇒ 12 + 2 d = 22
⇒ 2 d = 10 ⇒ d = 5
Putting value of d in equation a = 12 − 5 d,
a = 12 – 5 ( 5 ) = 12 – 25 = − 13
Therefore, first term = a = − 13 and, Common difference = d = 5
Therefore, AP is – 13, – 8, – 3, 2…
Its first three terms are – 13, – 8 and – 3.
- Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000.
Solution: Subba Rao’s starting salary = Rs 5000
It means, first term = a = 5000
He gets an increment of Rs 200 after every year.
Therefore, common difference = d = 200
His salary after 1 year = 5000 + 200 = Rs 5200
His salary after two years = 5200 + 200 = Rs 5400
Therefore, it is an AP of the form 5000, 5200, 5400, 5600… , 7000
We want to know in which year his income reaches Rs 7000.
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
7000 = 5000 + ( n − 1 ) ( 200 )
⇒ 7000 = 5000 + 200 n − 200
⇒ 7000 – 5000 + 200 = 200 n
⇒ 2200 = 200 n
⇒ n = 11
It means after 11 years, Subba Rao’s income would be Rs 7000.
- Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Solution: Ramkali saved Rs. 5 in the first week of year. It means first term =
a = 5
Ramkali increased her weekly savings by Rs 1.75.
Therefore, common difference = d = Rs 1.75
Money saved by Ramkali in the second week = a + d = 5 + 1.75 = Rs 6.75
Money saved by Ramkali in the third week = 6.75 + 1.75 = Rs 8.5
Therefore, it is an AP of the form:5, 6.75, 8.5 … , 20.75
We want to know in which year her weekly savings become 20.75.
Using formula an = a + ( n − 1 ) d, to find nth term of arithmetic progression,
20.75 = 5 + ( n − 1 ) ( 1.75 )
⇒ 20.75 = 5 + 1.75 n − 1.75
⇒ 17.5 = 1.75 n ⇒ n = 10
It means in the 10th week her savings become Rs 20.75.
