Get First Few Terms or Describe each Arithmetic Sequence, Given each of the Sums

Example 1: Find the First Three Terms
Find the first three terms of an arithmetic series in which a1=9, an=105, and Sn=741.
Solution steps:
Step 1. Since you know a1, an, and Sn, use Snn (a1+an) to find n.

Snn (a1+an)
741=½n (9+105)
741=57n
13n

Step 2. Find d.
an=a1+(n-1)d
105=9+(13-1)d
96=12d
8=d

Step 3. Use d to determine 112 and 113.
The first three terms are 9, 17, and 25.

Ex2: Determining the First Few Terms Given the Sum, Common Difference, and One Term
An arithmetic series has S20=143⅓, d=⅓ and t20=10⅓; determine the first 3 terms of the series.
solution:
S20 and t20 are known, so use this rule to determine t1:

Snn (t1+tn)

Substitute: n=20, S20=143⅓, t20=10⅓
143⅓=½⋅20⋅(t1+10⅓)

Simplify.
143⅓=10⋅(t1+10+⅓)
143⅓=10t1+100+3⅓
143⅓-3⅓-100=10t1

Solve for t1.
40=10t1
4=t1

The first term is 4 and the common dlfference is ⅓.
So, the first 3 terms of the series are written as the partial sum:
4+4⅓+4⅔

Let’s open a particular post in new tab Sum Formula for every Arithmetic Series with Given the First and Last Terms.
Ex3. The sum of three numbers in an A.P. is 12 and the sum of their cubes is 408. Find them.
Solution:
Let the required numbers be
ad, a, a+d

According to 1st condition:
(ad)+a+(a+d)=12
ad+a+a+d=12
3a=12
a=4

According to 2nd given condition:
(ad)3+a3+(a+d)3=408
a3d3-3a2d+ad2+a3+a3+d3+d3+3a2d+3ad2=408
3a3+6ad2=408
3(4)3+6(4) d2=408
24d2=408-192
d2=9
d=±3

When a=4, d=3 then number are ad, a, a+d
i.e. 4-3, 4 , 4+3 i.e. 1, 4, 7
when a=4, d=-3 then numbers are ad, a, a+d
4-(-3), 4, 4+(-3)
4+3, 4, 4-3 i.e., 7, 4, 1

Hence the required numbers are 1, 4, 7 or 7, 4, 1

Note:

The problem containing three or more numbers in A.P. whose sum in given it is often to assume the number as follows.

If the required numbers in A.P are odd i.e. 3, 5, 7 etc. Then take ‘a’ (first term) as the middle number and d as the common difference.

Thus three numbers are ad, a, a+d. If the required numbers in A.P are even i.e. 2, 4, 6, etc. then take ad, a+d as the middle numbers and 2d as the common difference.

Thus four numbers are a-3d, ad, a+d, a+3d and six numbers are:

a-5d, a-3d, ad, a+d, a+3d, a+5d etc.

Ex4. In an AP. It is given that S5+S7=167 and S10=235, then find the AP, where Sn denotes the sum of its first n terms.
Solution:
If you meet trouble to understand this solution, let’s open a particular post in new tab How to Form the Sum of the First n Terms of every Arithmetic Series?
Let a be the first term and d be the common difference of thee AP. Then,

S5+S7=167
Snn [2a+(n-1)d]
½⋅5⋅(2a+4d)+½⋅7⋅(2a+6d)=167
5a+10d+7a+21d=167
12a+31d=167… (1)

Also,
S10=235
½⋅10⋅(2a+9d)=235
5(2a+9d)=235
2a+9d=47

Multiplying both sides by 6, we get
12a+54d=282… (2)

Subtracting (1) from (2), we get
12a+54d-12a-31d=282-167
23d=115
d=5

Putting d=5 in (1), we get
12a+31⋅5=167
12a+155=167
12a=167-155=12
a=1.

Hence, the AP is 1, 6, 11, 16, … .

Ex5. The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1∶5, find the AP.
Solution:
Let a be the first term and d be the common difference of the AP.

S7=182
Snn [2a+(n-1)d]
½⋅7⋅(2a+6d)=182
a+3d=26 … (1)

Given a4=a17=1∶5.
[an=a+(n-1)d]
(a+3d)/(a+16d)=⅕
5a+15d=a+16d
d=4a … (2)

Solving (1) and (2), we get
a+3⋅4a=26
13a=26
a=2.

Putting a=2 in (2), we get
d=4⋅2=8.

Hence, the required AP is 2, 10, 18, 26, … .

Ex6. The sum first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.
Solution:
Let a be the first term and d be the common difference of the AP. Then, S10=-150 (Given).

Snn [2a+(n-1)d]
½⋅10⋅(2a+9d)=-150
5(2a+9d)=-150
2a+9d=-30 … (1)

It is given that the sum of its next 10 terms is -550.
Now, S20= Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms =
-150+(-550)=-700.
S20=-700
½⋅20⋅(2a+19d)=-700
10(2a+19d)=-700
2a+19d=-70… (2)

Subtracting (1) from (2), we get
(2a+19d)-(2a+9d)=-70-(-30)
10d=-40
d=-4

Putting d=-4 in (1), we get
2a+9⋅(-4)=-30
2a=-30+36=6
a=3

Hence, the required AP is 3, -1 -5, -9, … .

Ex7. A sum of $700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is $20 less than its preceding prize, find the value of each prize.
Solution:
Let the value of the first prize be a.
Since the value of each prize is 20 less than its preceding prize, so the values of the prizes are in AP with common difference $20.

½⋅40⋅[2a+(40-1)d]=36000
20(2a+39d)=36000
2a+39d=1800… (2)

Number of cash prizes to be given to the students, n=7.
Total sum of the prizes, S7=$700
Using the formula, Snn [2a+(n-1)d], we get
S7=½⋅7⋅[2a+(7-1)⋅(-20)]=700
½⋅7⋅(2a-120)=700
7a-420=700
7a=700+420=1120
a=160

Thus, the value of the first prize is $160.
Hence, the value of each prize is $160, $140, $120, $100, $80, $60 and $40.

Find the first three terms of each arithmetic series Ex8 to Ex9.
Ex8. a1=8, an=100, Sn=1296
Solution:

Snn ⋅(a1+an)
1296n⋅½⋅(8+100)
1296n⋅54
n=1296÷54=24

an=a1+(n-1)d
100=8+(24-1)d
23d=92
d=4

Therefore, the first three terms are 8, (8+4) or 12, (12+4) or 16.
Answer: 8, 12, 16

Ex9. n=18, an=112, Sn=1098
Solution:

Snn ⋅(a1+an)
1098=½⋅18⋅(a1+112)
a1+112=1098÷9
a1=122-112
a1=10

an=a1+(n-1)d
112=10+(18-1)d
17d=102
d=6

Therefore, the first three terms are 10, (10+6) or 16, (16+6) or 22.
Answer: 10, 16, 22

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