# Get the Number of Terms of each Arithmetic Sequence, Given each of the Sums

Problem 1. The sum of the first and the last terms of an arithmetic sequence is 80 and the sum of all the terms is 1, 200. How many terms are in the sequence?
Solution:

Snn(a1+an)
1, 200=½n(80)
1, 200=40n
30=n

Answer: There are 30 terms in the sequence.

P2. The first term of an arithmetic sequence is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Here, a=5, an=45 and Sn=400.
The sum of n terms of an arithmetic sequence is given by

Snn(a+an)
400=½n[5+45]
400=25n
n=16
an=a+(n-1)d
45=5+(16-1)d
40=15d
d=40/15 … (5/5)=8/3=2⅔

Hence, the number of terms are 16 and the common difference is 2⅔.

P3. The sum of the first four terms of an arithmetic sequence is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Solution:
Let the arithmetic sequence be a, a+d, a+2d, a+3d, …, a+(n-2)d, a+(n-1)d.
Sum of first four terms =a+(a+d)+(a+2d)+(a+3d)=4a+6d.
Sum of last four terms =[a+(n-4)d]+[a+(n-3)d]+[a+(n-2)d]+[a+(n-1)d]

=4a+(4n-10)d

According to the given condition,
4a+6d=56
4(11)+6d=56 [Since a=11 (given)]
6d=12
d=2
4a+(4n-10)d=112
4(11)+(4n-10)⋅2=112
(4n-10)⋅2=68
4n-10=34
4n=44
n=11

Thus, the number of terms of the arithmetic sequence is 11.

P4. How many terms of the arithmetic sequence: 24, 21, 18, … must be taken so that their sum is 78?
Solution:
Here, a=24, d=21-24=-3, Sn=78. We need to find n.
We know that Snn[2a+(n-1)d]. So

78=½n[2⋅24+(n-1)(-3)]
78⋅2=n[48-3n+3]
156=48n-3n2+3n
3n2-51n+156=0 … (÷3)
n2-17n+52=0
(n-4)(n-13)=0
n=4 or n=13

Both values of n are admissible. So, the number of terms is either 4 or 13.

Remarks:
(i). In this case, the sum of the first 4 terms=the sum of the first 13 terms=78.
(ii). Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.

P5. How many terms of the arithmetic sequence -6, -11/2, -5, … are needed to give the sum -25?
Solution:
Let the sum of n terms of the given arithmetic sequence be -25.
It is known that, Snn[2a+d(n-1)], where n= number of terms, a= first term, and d= common difference
Here, a=-6,

d=-11/2-(-6)=6-5½=½

Therefore, we obtain
-25=½n[2(-6)+½(n-1)] … (×2)
-50=n(-12+½n-½) … (×2)
-100=n(-24+n-1)
-100=n2-25n
0=n2-25n+100
0=(n-5)(n=-20)
n=5 or n=20.

P6. How many terms of the arithmetic sequence 63, 60, 57, 54, … must be taken so that their sum is 693? Explain the double answer.
Solution:
The given arithmetic sequence is 63, 60, 57, 54, … .
Here, a=63 and d=60-63=-3.
Let the required number of terms be n. Then, Sn=693.

½n⋅[2⋅63+(n-1)⋅(-3)]=693
½n⋅(126-3n+3)=693
n(129-3n)=1386
3n2-129n+1386=0
3n2-66n-63n+1386=0
3n(n-22)-63(n – 22)=0
(n-22)(3n-63)=0
n-22=0 or 3n-63=0
n=22 or n=21

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of the arithmetic sequence is 0.
a22=63+(22-1)⋅(-3)=63-63=0

Hence, the required number of terms is 21 or 22.

P7. How many terms of the arithmetic sequence 21, 18, 15, must be added to get the sum 0?
Solution
The given arithmetic sequence is 21, 18, 15, … .
Here, a=21 and d=18-21=-3.
Let the required number of terms be n. Then, Sn=0.

Snn[2a+(n-1)d]
½n⋅[2⋅21+(n-1)⋅(-3)]=0
½n⋅[42-3n+3]=0
n⋅(45-3n)=0
n=0 or 45-3n=0
n=0 or n=15
n=15 … (Number of terms cannot be zero in this sequence)

Hence, the required number of terms is 15.

P8. 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 (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row? Solution:
Number of logs in bottom row =20, logs in next row =19, logs in next row =18 Similarly, the series of number of logs is 20, 19, 18, 17, … .
Here, a=20, d=19-20=-1 and Sn=200.
The sum of n terms of an arithmetic sequence is given by Snn[2a+(n-1)d].
200=½¬n[2⋅20+(n-1)⋅(-1)]
400=n⋅(40-n+1)
400=41nn2
n2-41n+400=0
(n-16)(n-25)=0
n=16 or n=25

If n=16, a16=a+15d=20+15(-1)=5
If n=25, a25=a+24d=20+24(-1)=-4, which is not possible.
Hence, the 200 logs are placed in 16 rows and 5 logs are in the top row.

P9. The sum of the series 5+3+1+⋯ is -216, determine the number of terms in the series
Solution:
a=5, d=-2, Sn=-216, Snn[2a+(n-1)d], n?
Substitute into the formula:

-216=½n[2(5)+(n-1)(-2)]
-216=½n[10-2n+2]
-216=½n[12-2n]
-432=12n-2n2
-432=-2n2+12n … . Make equation =0
2n2-12n-432=0 … . Divide through by 2 (common factor)
n2-6n-216=0 … . Factorise trinomial
(n-18)(n+12)=0
n-18=0 or n+12=0
n=18 or n=-12
n>0 ∴n=18

∴ 18 terms of the series add up to -216.

P10. The first and second terms of an arithmetic sequence are 10 and 6 respectively.
a) Calculate the 11th term of the sequence.
b) The sum of the first n terms of this sequence is -560. Calculate n.
Solution:
a) an=a+(n-1)d

a11=10+(11-1)(-4)=-30

b) Snn[2a+(n-1)d]
-560=½n[2⋅10+(n-1)(-4)]
-1120=-4n2+24n
4n2-24n-1120=0
n2-6n-280=0
(n-20)(n+14)=0
n=20 or n=-14

n=20 only because number of terms cannot be a negative number.

P11. How many terms of the arithmetic sequence: 9, 17, 25 … must be taken to give a sum of 636?
Solution:
Here, a=9, d=17-9=8 and Sn=636. The sum of n terms of an arithmetic sequence is given by

Snn[2a+(n-1)d]
636=½n[2(9)+(n-1)⋅8]
636=n[9+4n-4]
4n2+5n-636=0
4n2+53n-48n-636=0
n(4n+53)-12(4n+53)=0
(n-12)(4n+53)=0
n-12=0
n=12
4n+53=0→n=-53/4, not taken due to (+)n.

Hence, 12 terms of the arithmetic sequence: 9, 17, 25 … must be taken to get the sum 636.

P12. How many terms of the arithmetic sequence 9, 17, 25, must be taken so that their sum is 636?
Solution:
The given arithmetic sequence is 9, I7, 25, … .
Here, a=9 and d=17-9=8.
Let the required number of terms be n. Then, Sn=636.

Snn[2a+(n-1)d]
½n⋅[2⋅9+(n-1)8]=636
½n⋅[18+8n-8]=636
½n⋅(10+8n)=636
n(4n+5)=636
4n2+5n-636=0
4n2-48n+53n-636=0
4n(n-2)+53(n-2)=0
(n-2)(4n+53)
n-2=0 or 4n+53=0
n=12 or n=(-53)/4
n=12 … (Number of terms cannot negative)

Hence, the required number of terms is 12.

P13. How many terms of the arithmetic sequence 3, 7, 11, 15, … will make the sum 406?
(a) 10 (b) 12 (c) 14 (d) 20
Solution:
Here, a=3 and d=(7-3)=4.
Let the sum of n terms be 406. Then, we have:

Snn[2a+(n-1)d]
½n[2⋅3+(n-1)⋅4]=406
n[3+2n-2]=406
2n2-28n+29n=406
2n2+n-406=0
2n2-28n+29n-406=0
2n(n-14)+29(n-14)=0
(2n+29)(n-14)=0
n=14 (∵ n can’t be a fraction).

Hence, 14 terms will make the sum 406.

P14. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Solution:
Let x be the number of days in which 150 workers finish the work.
According to the given information,

150x=150+146+142+⋯ (x+8) terms

The series 150+146+142+⋯ (x+8) terms is an arithmetic sequence with first term 146, common difference -4 and number of terms as (x+8).
⇒150x=½(x+8)[2⋅150+(x+8-1)(-4)]
300x=(x+8)[300-4x-28]
300x=(x+8)(272-4x)
300x=272x-4x2+2176-32x
4x2+300x-240x-2176=0
4x2+60x-2176=0 … (÷4)
x2+15x-544=0 … (÷4)
(x-17)(x+32)=0
x=17 or x=-32

However, x cannot be negative.
x=17.
Therefore, originally, the number of days in which the work was completed is 17.
Thus, required number of days =(17+8)=25.

P15. In an arithmetic sequence: given a=2, d=8, Sn=90, find n and an.
Here, a=2, d=8 and Sn=90.
The sum of n terms of an arithmetic sequence is given by P16. The sum of the series 22+28+34+ … is 1870. Determine the number of terms.
Solution:
Here, a=22 and d=6. n cannot be a negative because it is the number of terms.

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