ARITHMETIC AND GEOMETRIC PROGRESSIONS 53

A

125 − 80 = 5 d^2

45 = 5 d^2

from which,d^2 =

45

5

=9. Henced=

√

9 =±3.

The three numbers are thus (5−3), 5 and (5+3),
i.e.2, 5 and 8.

Problem 8. Find the sum of all the numbers

between 0 and 207 which are exactly divisible

by 3.

The series 3, 6, 9, 12,..., 207 is an AP whose first
terma=3 and common differenced= 3

The last term is a+(n−1)d= 207

i.e. 3 +(n−1)3=207,

from which (n−1)=

207 − 3

3

= 68

Hence n= 68 + 1 = 69

The sum of all 69 terms is given by

S 69 =

n

2

[2a+(n−1)d]

=

69

2

[2(3)+(69−1)3]

=

69

2

[6+204]=

69

2

(210)= 7245

Problem 9. The first, twelfth and last term of

an arithmetic progression are 4, 31^12 , and 376^12

respectively. Determine (a) the number of terms

in the series, (b) the sum of all the terms and

(c) the ‘80’th term.

(a) Let the AP bea,a+d,a+ 2 d,...,a+(n−1)d,

wherea= 4

The 12th term is:a+(12−1)d= (^3112)
i.e. 4 + 11 d= 3112 ,
from which, 11d= 3112 − 4 = (^2712)
Henced=
(^2712)
11
= (^212)
The last term isa+(n−1)d
i.e. 4+(n−1)
(
(^212)
)
= (^37612)
(n−1)=
37612 − 4
(^212)

(^37212)
(^212)
= 149
Hence the number of terms in the series,
n= 149 + 1 = 150
(b) Sum of all the terms,
S 150 =
n
2
[2a+(n−1)d]

150
2
[
2(4)+(150−1)
(
2
1
2
)]
= 75
[
8 +(149)
(
2
1
2
)]
=85[8+ 372 .5]
=75(380.5)= 28537
1
2
(c) The 80th term is:
a+(n−1)d= 4 +(80−1)
(
(^212)
)
= 4 +(79)
(
(^212)
)
= 4 + 197. 5 = (^20112)
Now try the following exercise.
Exercise 29 Further problems on arith-
metic progressions

1. The sum of 15 terms of an arithmetic progres-
   sion is 202.5 and the common difference is 2.
   Find the first term of the series. [−0.5]


2. Three numbers are in arithmetic progression.
   Their sum is 9 and their product is 20.25.
   Determine the three numbers. [1.5, 3, 4.5]


3. Find the sum of all the numbers between 5
   and 250 which are exactly divisible by 4.
   [7808]


4. Find the number of terms of the series 5, 8,
   11,...of which the sum is 1025. [25]